Ramanujan's radical and how we define an infinite nested radical












9












$begingroup$


I know it is true that we have



$$sqrt{1+2sqrt{1+3sqrt{1+4sqrt{1+cdots}}}}=3$$



The argument is to break the nested radical into something like



$3 = sqrt{9}=sqrt{1+2sqrt{16}}=sqrt{1+2sqrt{1+3sqrt{25}}}=...=$$sqrt{1+2sqrt{1+3sqrt{1+4sqrt{1+cdots}}}}$



However, I am not convinced. I can do something like



$4 = sqrt{16}=sqrt{1+2sqrt{56.25}}=sqrt{1+2sqrt{1+3sqrt{frac{48841}{144}}}}=...=$$sqrt{1+2sqrt{1+3sqrt{1+4sqrt{1+cdots}}}}$



Something must be wrong and and the reason behind should be misunderstanding of how we define infinte nested radical in the form of
$ sqrt{a_{0}+a_{1}sqrt{a_{2}+a_{3}sqrt{a_{4}+a_{5}sqrt{a_{6}+cdots}}}} $
. I researched for a while but all I could find was computation tricks but not strict definition. Really need help here. Thanks






real-analysis sequences-and-series elementary-number-theory







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Surely the rigorous definition is that it is the limit of the finite expressions that you get when you drop the $+ cdots$, provided that this limit exists.
    $endgroup$
    – John Coleman
    4 hours ago












  • $begingroup$
    So you are suggesting $sqrt{1+2sqrt{1+3sqrt{1+4sqrt{1+cdots}}}} $ actually has no universally-agreed definition in Maths?
    $endgroup$
    – Anson NG
    3 hours ago






  • 2




    $begingroup$
    When you write something down, you stop your writing at some stage and break into $+ cdots$. What you have written is a purely finite expression. Think of these finite expressions as terms in a sequence. What do these terms approach?
    $endgroup$
    – John Coleman
    3 hours ago






  • 1




    $begingroup$
    I understand your comment. Given a general term of partial sum, we can find it limit although it might not exist. We can do it because we all universally agree what it means by a general term. Sometimes the dots is clear in the sense that we would not argue its meaning. For example,comsider $ 1+2+3+4+5+...$. Someone could still argue that the 6th term can be any number and does not have to be 6 but it's kind of agreed that we will look at the pattern and take the 6th term as 6.
    $endgroup$
    – Anson NG
    3 hours ago








  • 1




    $begingroup$
    This is interesting since you can redefine any number this way as you pointed out. We need to think of infinite nested root with more careful definition.. Maybe it's only equal to 3 if we don't have to use fraction to expand out more nested root. That is, in the Ramanujan's sum, when you expand the nested root, they are all in term of integers. Where as, for other number, you will get fraction like term. Similar to what you have... just a thought!
    $endgroup$
    – user209663
    2 hours ago


















9












$begingroup$


I know it is true that we have



$$sqrt{1+2sqrt{1+3sqrt{1+4sqrt{1+cdots}}}}=3$$



The argument is to break the nested radical into something like



$3 = sqrt{9}=sqrt{1+2sqrt{16}}=sqrt{1+2sqrt{1+3sqrt{25}}}=...=$$sqrt{1+2sqrt{1+3sqrt{1+4sqrt{1+cdots}}}}$



However, I am not convinced. I can do something like



$4 = sqrt{16}=sqrt{1+2sqrt{56.25}}=sqrt{1+2sqrt{1+3sqrt{frac{48841}{144}}}}=...=$$sqrt{1+2sqrt{1+3sqrt{1+4sqrt{1+cdots}}}}$



Something must be wrong and and the reason behind should be misunderstanding of how we define infinte nested radical in the form of
$ sqrt{a_{0}+a_{1}sqrt{a_{2}+a_{3}sqrt{a_{4}+a_{5}sqrt{a_{6}+cdots}}}} $
. I researched for a while but all I could find was computation tricks but not strict definition. Really need help here. Thanks






real-analysis sequences-and-series elementary-number-theory







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Surely the rigorous definition is that it is the limit of the finite expressions that you get when you drop the $+ cdots$, provided that this limit exists.
    $endgroup$
    – John Coleman
    4 hours ago












  • $begingroup$
    So you are suggesting $sqrt{1+2sqrt{1+3sqrt{1+4sqrt{1+cdots}}}} $ actually has no universally-agreed definition in Maths?
    $endgroup$
    – Anson NG
    3 hours ago






  • 2




    $begingroup$
    When you write something down, you stop your writing at some stage and break into $+ cdots$. What you have written is a purely finite expression. Think of these finite expressions as terms in a sequence. What do these terms approach?
    $endgroup$
    – John Coleman
    3 hours ago






  • 1




    $begingroup$
    I understand your comment. Given a general term of partial sum, we can find it limit although it might not exist. We can do it because we all universally agree what it means by a general term. Sometimes the dots is clear in the sense that we would not argue its meaning. For example,comsider $ 1+2+3+4+5+...$. Someone could still argue that the 6th term can be any number and does not have to be 6 but it's kind of agreed that we will look at the pattern and take the 6th term as 6.
    $endgroup$
    – Anson NG
    3 hours ago








  • 1




    $begingroup$
    This is interesting since you can redefine any number this way as you pointed out. We need to think of infinite nested root with more careful definition.. Maybe it's only equal to 3 if we don't have to use fraction to expand out more nested root. That is, in the Ramanujan's sum, when you expand the nested root, they are all in term of integers. Where as, for other number, you will get fraction like term. Similar to what you have... just a thought!
    $endgroup$
    – user209663
    2 hours ago
















9












9








9


4



$begingroup$


I know it is true that we have



$$sqrt{1+2sqrt{1+3sqrt{1+4sqrt{1+cdots}}}}=3$$



The argument is to break the nested radical into something like



$3 = sqrt{9}=sqrt{1+2sqrt{16}}=sqrt{1+2sqrt{1+3sqrt{25}}}=...=$$sqrt{1+2sqrt{1+3sqrt{1+4sqrt{1+cdots}}}}$



However, I am not convinced. I can do something like



$4 = sqrt{16}=sqrt{1+2sqrt{56.25}}=sqrt{1+2sqrt{1+3sqrt{frac{48841}{144}}}}=...=$$sqrt{1+2sqrt{1+3sqrt{1+4sqrt{1+cdots}}}}$



Something must be wrong and and the reason behind should be misunderstanding of how we define infinte nested radical in the form of
$ sqrt{a_{0}+a_{1}sqrt{a_{2}+a_{3}sqrt{a_{4}+a_{5}sqrt{a_{6}+cdots}}}} $
. I researched for a while but all I could find was computation tricks but not strict definition. Really need help here. Thanks






real-analysis sequences-and-series elementary-number-theory







share|cite|improve this question











$endgroup$




I know it is true that we have



$$sqrt{1+2sqrt{1+3sqrt{1+4sqrt{1+cdots}}}}=3$$



The argument is to break the nested radical into something like



$3 = sqrt{9}=sqrt{1+2sqrt{16}}=sqrt{1+2sqrt{1+3sqrt{25}}}=...=$$sqrt{1+2sqrt{1+3sqrt{1+4sqrt{1+cdots}}}}$



However, I am not convinced. I can do something like



$4 = sqrt{16}=sqrt{1+2sqrt{56.25}}=sqrt{1+2sqrt{1+3sqrt{frac{48841}{144}}}}=...=$$sqrt{1+2sqrt{1+3sqrt{1+4sqrt{1+cdots}}}}$



Something must be wrong and and the reason behind should be misunderstanding of how we define infinte nested radical in the form of
$ sqrt{a_{0}+a_{1}sqrt{a_{2}+a_{3}sqrt{a_{4}+a_{5}sqrt{a_{6}+cdots}}}} $
. I researched for a while but all I could find was computation tricks but not strict definition. Really need help here. Thanks






real-analysis sequences-and-series elementary-number-theory




real-analysis sequences-and-series elementary-number-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 4 hours ago









user328442

1,8551516




1,8551516










asked 4 hours ago









Anson NGAnson NG

676




676








  • 1




    $begingroup$
    Surely the rigorous definition is that it is the limit of the finite expressions that you get when you drop the $+ cdots$, provided that this limit exists.
    $endgroup$
    – John Coleman
    4 hours ago












  • $begingroup$
    So you are suggesting $sqrt{1+2sqrt{1+3sqrt{1+4sqrt{1+cdots}}}} $ actually has no universally-agreed definition in Maths?
    $endgroup$
    – Anson NG
    3 hours ago






  • 2




    $begingroup$
    When you write something down, you stop your writing at some stage and break into $+ cdots$. What you have written is a purely finite expression. Think of these finite expressions as terms in a sequence. What do these terms approach?
    $endgroup$
    – John Coleman
    3 hours ago






  • 1




    $begingroup$
    I understand your comment. Given a general term of partial sum, we can find it limit although it might not exist. We can do it because we all universally agree what it means by a general term. Sometimes the dots is clear in the sense that we would not argue its meaning. For example,comsider $ 1+2+3+4+5+...$. Someone could still argue that the 6th term can be any number and does not have to be 6 but it's kind of agreed that we will look at the pattern and take the 6th term as 6.
    $endgroup$
    – Anson NG
    3 hours ago








  • 1




    $begingroup$
    This is interesting since you can redefine any number this way as you pointed out. We need to think of infinite nested root with more careful definition.. Maybe it's only equal to 3 if we don't have to use fraction to expand out more nested root. That is, in the Ramanujan's sum, when you expand the nested root, they are all in term of integers. Where as, for other number, you will get fraction like term. Similar to what you have... just a thought!
    $endgroup$
    – user209663
    2 hours ago
















  • 1




    $begingroup$
    Surely the rigorous definition is that it is the limit of the finite expressions that you get when you drop the $+ cdots$, provided that this limit exists.
    $endgroup$
    – John Coleman
    4 hours ago












  • $begingroup$
    So you are suggesting $sqrt{1+2sqrt{1+3sqrt{1+4sqrt{1+cdots}}}} $ actually has no universally-agreed definition in Maths?
    $endgroup$
    – Anson NG
    3 hours ago






  • 2




    $begingroup$
    When you write something down, you stop your writing at some stage and break into $+ cdots$. What you have written is a purely finite expression. Think of these finite expressions as terms in a sequence. What do these terms approach?
    $endgroup$
    – John Coleman
    3 hours ago






  • 1




    $begingroup$
    I understand your comment. Given a general term of partial sum, we can find it limit although it might not exist. We can do it because we all universally agree what it means by a general term. Sometimes the dots is clear in the sense that we would not argue its meaning. For example,comsider $ 1+2+3+4+5+...$. Someone could still argue that the 6th term can be any number and does not have to be 6 but it's kind of agreed that we will look at the pattern and take the 6th term as 6.
    $endgroup$
    – Anson NG
    3 hours ago








  • 1




    $begingroup$
    This is interesting since you can redefine any number this way as you pointed out. We need to think of infinite nested root with more careful definition.. Maybe it's only equal to 3 if we don't have to use fraction to expand out more nested root. That is, in the Ramanujan's sum, when you expand the nested root, they are all in term of integers. Where as, for other number, you will get fraction like term. Similar to what you have... just a thought!
    $endgroup$
    – user209663
    2 hours ago










1




1




$begingroup$
Surely the rigorous definition is that it is the limit of the finite expressions that you get when you drop the $+ cdots$, provided that this limit exists.
$endgroup$
– John Coleman
4 hours ago






$begingroup$
Surely the rigorous definition is that it is the limit of the finite expressions that you get when you drop the $+ cdots$, provided that this limit exists.
$endgroup$
– John Coleman
4 hours ago














$begingroup$
So you are suggesting $sqrt{1+2sqrt{1+3sqrt{1+4sqrt{1+cdots}}}} $ actually has no universally-agreed definition in Maths?
$endgroup$
– Anson NG
3 hours ago




$begingroup$
So you are suggesting $sqrt{1+2sqrt{1+3sqrt{1+4sqrt{1+cdots}}}} $ actually has no universally-agreed definition in Maths?
$endgroup$
– Anson NG
3 hours ago




2




2




$begingroup$
When you write something down, you stop your writing at some stage and break into $+ cdots$. What you have written is a purely finite expression. Think of these finite expressions as terms in a sequence. What do these terms approach?
$endgroup$
– John Coleman
3 hours ago




$begingroup$
When you write something down, you stop your writing at some stage and break into $+ cdots$. What you have written is a purely finite expression. Think of these finite expressions as terms in a sequence. What do these terms approach?
$endgroup$
– John Coleman
3 hours ago




1




1




$begingroup$
I understand your comment. Given a general term of partial sum, we can find it limit although it might not exist. We can do it because we all universally agree what it means by a general term. Sometimes the dots is clear in the sense that we would not argue its meaning. For example,comsider $ 1+2+3+4+5+...$. Someone could still argue that the 6th term can be any number and does not have to be 6 but it's kind of agreed that we will look at the pattern and take the 6th term as 6.
$endgroup$
– Anson NG
3 hours ago






$begingroup$
I understand your comment. Given a general term of partial sum, we can find it limit although it might not exist. We can do it because we all universally agree what it means by a general term. Sometimes the dots is clear in the sense that we would not argue its meaning. For example,comsider $ 1+2+3+4+5+...$. Someone could still argue that the 6th term can be any number and does not have to be 6 but it's kind of agreed that we will look at the pattern and take the 6th term as 6.
$endgroup$
– Anson NG
3 hours ago






1




1




$begingroup$
This is interesting since you can redefine any number this way as you pointed out. We need to think of infinite nested root with more careful definition.. Maybe it's only equal to 3 if we don't have to use fraction to expand out more nested root. That is, in the Ramanujan's sum, when you expand the nested root, they are all in term of integers. Where as, for other number, you will get fraction like term. Similar to what you have... just a thought!
$endgroup$
– user209663
2 hours ago






$begingroup$
This is interesting since you can redefine any number this way as you pointed out. We need to think of infinite nested root with more careful definition.. Maybe it's only equal to 3 if we don't have to use fraction to expand out more nested root. That is, in the Ramanujan's sum, when you expand the nested root, they are all in term of integers. Where as, for other number, you will get fraction like term. Similar to what you have... just a thought!
$endgroup$
– user209663
2 hours ago












2 Answers
2






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Introduction:



The issue is what "..." really "represents."



Typically we use it as a sort of shorthand, as if to say "look, I can't write infinitely many things down, just assume that the obvious pattern holds and goes on infinitely."



This idea holds for all sorts of things - nested radicals, infinite sums, continued fractions, infinite sequences, etc.




On Infinite Sums:



A simple example: the sum of the reciprocals of squares:



$$1 + frac{1}{4} + frac{1}{9} + frac{1}{16} + ...$$



This is a well known summation. It is the Riemann zeta function $zeta(s)$ at $s=2$, and is known to evaluate to $pi^2/6$ (proved by Euler and known as the Basel problem).



Another, easier-to-handle summation is the geometric sum



$$1 + frac 1 2 + frac 1 4 + frac 1 8 + ...$$



This is a geometric series where the ratio is $1/2$ - each summand is half the previous one. We know, too, that this evaluates to $2$.



Another geometric series you might see in proofs that $0.999... = 1$ is



$$9 left( frac{1}{10} + frac{1}{100} + frac{1}{1,000} + frac{1}{10,000} + ... right)$$



which equals $1$. In fact, any infinite geometric series, with first term $a$ and ratio $|r|<1$ can be evaluated by



$$sum_{n=0}^infty ar^n = frac{a}{1-r}$$



So a question arises - ignoring these "obvious" results (depending on your amount of mathematical knowledge), how would we know these converge to the given values? What, exactly, does it mean for a summation to converge to a number or equal a number? For finite sums this is no issue - if nothing else, we could add up each number manually, but we can't just add up every number from a set of infinitely-many numbers.



Well, one could argue by common sense that, if the sequence converges to some number, the more and more terms you add up, the closer they'll get to that number.



So we obtain one definition for the convergence of an infinite sum. Consider a sequence where the $n^{th}$ term is defined by the sum of the first $n$ terms in the sequence. To introduce some symbols, suppose we're trying to find the sum



$$sum_{k=1}^infty x_k = x_1 + x_2 + x_3 + x_4 + ...$$



for whatever these $x_i$'s are. Then define these so-called "partial sums" of this by a function $S(n)$:



$$S(n) = sum_{k=1}^n x_k = x_1 + x_2 + ... + x_n$$



Then we get a sequence of sums:



$$S_1, S_2, S_3, ...$$



or equivalently



$$x_1 ;;,;; x_1 + x_2;;,;; x_1 + x_2 + x_3;;,;; ...$$



Then we ask: what does $S(n)$ approach as $n$ grows without bound, if anything at all? (In calculus, we call this "the limit of the partial sums $S(n)$ as $n$ approaches infinity.")



For the case of our first geometric sum, we immediately see the sequence of partial sums



$$1, frac{3}{2}, frac{5}{4}, frac{15}{8},...$$



Clearly, this suggests a pattern - and if you want to, you can go ahead and prove it, I won't do so here for brevity's sake. The pattern is that the $n^{th}$ term of the sequence is



$$S(n) = frac{2n-1}{n}$$



We can then easily consider the limit of these partial sums:



$$lim_{ntoinfty} S(n) = lim_{ntoinfty} frac{2n-1}{n} = lim_{ntoinfty} 2 - frac 1 n = lim_{ntoinfty} 2 - lim_{ntoinfty} frac 1 n$$



Obviously, $1/n to 0$ as $n$ grows without bound, and $2$ is not affected by $n$, so we conclude



$$lim_{ntoinfty} S(n) = lim_{ntoinfty} 2 - lim_{ntoinfty} frac 1 n = 2 - 0 = 2$$



And thus we say



$$sum_{k=0}^infty left(frac 1 2 right)^k = 1 + frac 1 2 + frac 1 4 + frac 1 8 + ... = 2$$



because the partial sums approach $2$.




On Continued Fractions:



That was a simple, "first" sort of example, but mathematicians essentially do the same thing in other contexts. I want to touch on one more such context before we deal with the radical case, just to nail that point home.



In this case, it will be with continued fractions. One of the simpler such fractions is the one for $1$:



$$1 = frac{1}{2-frac{1}{2-frac{1}{2-frac{1}{...}}}}$$



As usual, the "..." denotes that this continues forever. But what it does it mean for this infinite expression to equal $1$?



For this, we consider a more general analogue of the "partial sum" from before - a "convergent." We cut up the sequence at logical finite points, whatever those points being depending on the context. Then if the sequence of the convergents approaches a limit, we say they're equal.



What are the convergents for a continued fraction? By convention, we cut off just before the start of the next fraction. That is, in the continued fraction for $1$, we cut off at the $n^{th} 2$ for the $n^{th}$ convergent, and ignore what follows. So we get the sequence of convergents



$$frac{1}{2} , frac{1}{2-frac{1}{2}}, frac{1}{2-frac{1}{2-frac{1}{2}}}$$



Working out the numbers, we find the sequence to be



$$frac{1}{2},frac{2}{3},frac{3}{4},...$$



Again, we see a pattern! The $n^{th}$ term of the sequence is clearly of the form



$$frac{n-1}{n}$$



Let $C(n)$ be a function denoting the $n^{th}$ convergent. Then $C(1)=1/2,$ $C(2) = 2/3,$ $C(n)=(n-1)/n,$ and so on. So like before we consider the infinite limit:



$$lim_{ntoinfty} C(n) = lim_{ntoinfty} frac{n-1}{n} = lim_{ntoinfty} 1 - frac 1 n = lim_{ntoinfty} 1 - lim_{ntoinfty} frac 1 n = 1 - 0 = 1$$



Thus we can conclude that the continued fraction equals $1$, because its sequence of convergents equals $1$!




On Infinite Radicals:



So now, we touch on infinite nested radicals. They're messier to deal with but doable.



One of the simpler examples of such radicals to contend with is



$$2 = sqrt{2 +sqrt{2 +sqrt{2 +sqrt{2 +sqrt{2 +...}}}}}$$



As with the previous two cases we see an infinite expression. We instinctively conclude by now: to logically define a limit for this expression - to assign it a value provided it even exists - we need to chop this up at finite points, defining a sequence of convergents $C(n)$, and then find $C(n)$ as $ntoinfty$.



Nested radicals are a lot messier than the previous, but we manage.



So first let the sequence of convergents be given by cutting off everything after the $n^{th} 2$ in the expression. Thus we get the sequence



$$sqrt 2 ;;,;; sqrt{2 + sqrt{2}};;,;; sqrt{2+sqrt{2+sqrt{2}}};;,;; sqrt{2+sqrt{2+sqrt{2+sqrt{2}}}}$$



Okay this isn't particularly nice already, but apparently there does exist, shockingly enough, a closed-form explicit expression for $C(n)$: (from: S. Zimmerman, C. Ho)



$$C(n) = 2cosleft(frac{pi}{2^{n+1}}right)$$



(I had to find that expression by Googling, I honestly didn't know that offhand. It can be proved by induction, as touched on in this MSE question.)



So luckily, then, we can find the limit of $C(n)$:



$$lim_{ntoinfty} C(n) = lim_{ntoinfty} 2cosleft(frac{pi}{2^{n+1}}right)$$



It is probably obvious that the argument of the cosine function approaches $0$ as $n$ grows without bound, and thus



$$lim_{ntoinfty} C(n) = lim_{ntoinfty} 2cosleft(frac{pi}{2^{n+1}}right) = lim_{ntoinfty} 2cos(0) = 2cdot 1 = 2$$



Thus, since its convergents approach $2$, we can conclude that



$$2 = sqrt{2 +sqrt{2 +sqrt{2 +sqrt{2 +sqrt{2 +...}}}}}$$




A Lengthy Conclusion:



So, in short, how do we evaluation an infinite expression, be it radical, continued fraction, sum, or otherwise?



We begin by truncating the expression at convenient finite places, creating a series of convergents, generalizations of the "partial sums" introduced in calculus. We then try to get a closed form or some other usable expression for the convergents $C(n)$, and consider the value as $ntoinfty$. If it converges to some value, we say that the expression is in fact equal to that value. If it doesn't, then the expression doesn't converge to any value.



This doesn't mean each expression is "nice." Radical expressions in particular, in my experience, tend to be nasty as all hell, and I'm lucky I found that one closed form expression for the particularly easy radical I chose.



This doesn't mean that other methods cannot be used to find the values, so long as there's some sort of logical justification for said method. For example, there is a justification for the formula for an infinite (and finite) geometric sum. We might have to circumvent the notion of partial sums entirely, or at least it might be convenient to do so. For example, with the Basel problem, Euler's proof focused on Maclaurin series, and none of this "convergent" stuff. (That proof is here plus other proofs of it!)



Luckily, at least, this notion of convergents, even if it may not always be the best way to do it, lends itself to an easy way to check a solution to any such problem. Just find a bunch of the convergents - take as many as you need. If you somehow have multiple solutions, as you think with Ramanujan's radical, then you'll see the convergents get closer and closer to the "true" solution. (How many convergents you need to find depends on the situation and how close your proposed solutions are. It might be immediately obvious after $10$ iterations, or might not be until $10,000,000$.)



So I will conclude by showing, in this way, that the solution is $3$ to Ramanujan's radical.



We begin with the radical itself:



$$sqrt{1+2sqrt{1+3sqrt{1+4sqrt{1+cdots}}}}=3$$



Let us begin by getting a series of convergents:



$$sqrt{1} ;;,;; sqrt{1 + 2sqrt{1}} ;;,;; sqrt{1 + 2sqrt{1 + 3sqrt{1}}} ;;,;;$$



Because the $sqrt{1}$ isn't necessary, we just let it be $1$.



$$1 ;;,;; sqrt{1 + 2} ;;,;; sqrt{1 + 2sqrt{1 + 3}} ;;,;;$$



Okay so ... where to go from here? Honestly, my initial temptation was to just use a MATLAB script and evaluate it, but I can't think of even a recursive closed form for this that would be nice enough. So in any event, we just have to go by "hand" (and by hand I mean WolframAlpha). Let $C(n)$ be the $n^{th}$ convergent. Then




  • $C(1) = 1$

  • $C(2) approx 1.732$

  • $C(3) approx 2.236$

  • $C(4) approx 2.560$

  • $C(5) approx 2.755$

  • $C(6) approx 2.867$

  • $C(7) approx 2.929$

  • $C(8) approx 2.963$

  • $C(9) approx 2.981$

  • $C(10) approx 2.990$


To skip a few values because at this point the changes get minimal, I used a macro to make a quick code for $C(50)$ so I could put it into Microsoft Excel and got the approximate result



$$C(50) approx 2.999 ; 999 ; 999 ; 999 ; 99$$



So while not the most rigorous result, we can at least on an intuitive level feel like the convergents from Ramanujan's radical converge to $3$, not $4$ or any other number. Neglecting that this is not an ironclad proof of the convergence, at least intuitively then we can feel like



$$3 = sqrt{1+2sqrt{1+3sqrt{1+4sqrt{1+cdots}}}}$$




Whew! Hopefully that lengthy post was helpful to you!




A late footnote, but Mathologer on YouTube did a video on this very topic, so his video would give a decent summary of all this stuff as well. Here's a link.






share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    $4=sqrt{16}=sqrt{1+3sqrt{25}}=sqrt{1+3sqrt{1+4sqrt{36}}}=sqrt{1+3sqrt{1+4sqrt{1+5sqrt{49}}}}=sqrt{1+3sqrt{1+4sqrt{1+5sqrt{1+6sqrt{64}}}}}=sqrt{1+3sqrt{1+4sqrt{1+5sqrt{1+6sqrt{1+7sqrt{81}}}}}}$



    $=sqrt{1+3sqrt{1+4sqrt{1+5sqrt{1+6sqrt{1+7sqrt{1+8sqrt{1+cdots}}}}}}}$



    $5=sqrt{25}=sqrt{1+4sqrt{36}}=sqrt{1+4sqrt{1+5sqrt{49}}}=sqrt{1+4sqrt{1+5sqrt{1+6sqrt{64}}}}=sqrt{1+4sqrt{1+5sqrt{1+6sqrt{1+7sqrt{81}}}}}=sqrt{1+4sqrt{1+5sqrt{1+6sqrt{1+7sqrt{1+8sqrt{100}}}}}}$



    $=sqrt{1+4sqrt{1+5sqrt{1+6sqrt{1+7sqrt{1+8sqrt{1+9sqrt{1+cdots}}}}}}}$



    $vdots$



    $n=sqrt{1+(n-1)sqrt{1+nsqrt{1+(n+1)sqrt{1+(n+2)sqrt{1+(n+3)sqrt{1+(n+4)sqrt{1+cdots}}}}}}}$






    share|cite|improve this answer











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      $begingroup$

      Introduction:



      The issue is what "..." really "represents."



      Typically we use it as a sort of shorthand, as if to say "look, I can't write infinitely many things down, just assume that the obvious pattern holds and goes on infinitely."



      This idea holds for all sorts of things - nested radicals, infinite sums, continued fractions, infinite sequences, etc.




      On Infinite Sums:



      A simple example: the sum of the reciprocals of squares:



      $$1 + frac{1}{4} + frac{1}{9} + frac{1}{16} + ...$$



      This is a well known summation. It is the Riemann zeta function $zeta(s)$ at $s=2$, and is known to evaluate to $pi^2/6$ (proved by Euler and known as the Basel problem).



      Another, easier-to-handle summation is the geometric sum



      $$1 + frac 1 2 + frac 1 4 + frac 1 8 + ...$$



      This is a geometric series where the ratio is $1/2$ - each summand is half the previous one. We know, too, that this evaluates to $2$.



      Another geometric series you might see in proofs that $0.999... = 1$ is



      $$9 left( frac{1}{10} + frac{1}{100} + frac{1}{1,000} + frac{1}{10,000} + ... right)$$



      which equals $1$. In fact, any infinite geometric series, with first term $a$ and ratio $|r|<1$ can be evaluated by



      $$sum_{n=0}^infty ar^n = frac{a}{1-r}$$



      So a question arises - ignoring these "obvious" results (depending on your amount of mathematical knowledge), how would we know these converge to the given values? What, exactly, does it mean for a summation to converge to a number or equal a number? For finite sums this is no issue - if nothing else, we could add up each number manually, but we can't just add up every number from a set of infinitely-many numbers.



      Well, one could argue by common sense that, if the sequence converges to some number, the more and more terms you add up, the closer they'll get to that number.



      So we obtain one definition for the convergence of an infinite sum. Consider a sequence where the $n^{th}$ term is defined by the sum of the first $n$ terms in the sequence. To introduce some symbols, suppose we're trying to find the sum



      $$sum_{k=1}^infty x_k = x_1 + x_2 + x_3 + x_4 + ...$$



      for whatever these $x_i$'s are. Then define these so-called "partial sums" of this by a function $S(n)$:



      $$S(n) = sum_{k=1}^n x_k = x_1 + x_2 + ... + x_n$$



      Then we get a sequence of sums:



      $$S_1, S_2, S_3, ...$$



      or equivalently



      $$x_1 ;;,;; x_1 + x_2;;,;; x_1 + x_2 + x_3;;,;; ...$$



      Then we ask: what does $S(n)$ approach as $n$ grows without bound, if anything at all? (In calculus, we call this "the limit of the partial sums $S(n)$ as $n$ approaches infinity.")



      For the case of our first geometric sum, we immediately see the sequence of partial sums



      $$1, frac{3}{2}, frac{5}{4}, frac{15}{8},...$$



      Clearly, this suggests a pattern - and if you want to, you can go ahead and prove it, I won't do so here for brevity's sake. The pattern is that the $n^{th}$ term of the sequence is



      $$S(n) = frac{2n-1}{n}$$



      We can then easily consider the limit of these partial sums:



      $$lim_{ntoinfty} S(n) = lim_{ntoinfty} frac{2n-1}{n} = lim_{ntoinfty} 2 - frac 1 n = lim_{ntoinfty} 2 - lim_{ntoinfty} frac 1 n$$



      Obviously, $1/n to 0$ as $n$ grows without bound, and $2$ is not affected by $n$, so we conclude



      $$lim_{ntoinfty} S(n) = lim_{ntoinfty} 2 - lim_{ntoinfty} frac 1 n = 2 - 0 = 2$$



      And thus we say



      $$sum_{k=0}^infty left(frac 1 2 right)^k = 1 + frac 1 2 + frac 1 4 + frac 1 8 + ... = 2$$



      because the partial sums approach $2$.




      On Continued Fractions:



      That was a simple, "first" sort of example, but mathematicians essentially do the same thing in other contexts. I want to touch on one more such context before we deal with the radical case, just to nail that point home.



      In this case, it will be with continued fractions. One of the simpler such fractions is the one for $1$:



      $$1 = frac{1}{2-frac{1}{2-frac{1}{2-frac{1}{...}}}}$$



      As usual, the "..." denotes that this continues forever. But what it does it mean for this infinite expression to equal $1$?



      For this, we consider a more general analogue of the "partial sum" from before - a "convergent." We cut up the sequence at logical finite points, whatever those points being depending on the context. Then if the sequence of the convergents approaches a limit, we say they're equal.



      What are the convergents for a continued fraction? By convention, we cut off just before the start of the next fraction. That is, in the continued fraction for $1$, we cut off at the $n^{th} 2$ for the $n^{th}$ convergent, and ignore what follows. So we get the sequence of convergents



      $$frac{1}{2} , frac{1}{2-frac{1}{2}}, frac{1}{2-frac{1}{2-frac{1}{2}}}$$



      Working out the numbers, we find the sequence to be



      $$frac{1}{2},frac{2}{3},frac{3}{4},...$$



      Again, we see a pattern! The $n^{th}$ term of the sequence is clearly of the form



      $$frac{n-1}{n}$$



      Let $C(n)$ be a function denoting the $n^{th}$ convergent. Then $C(1)=1/2,$ $C(2) = 2/3,$ $C(n)=(n-1)/n,$ and so on. So like before we consider the infinite limit:



      $$lim_{ntoinfty} C(n) = lim_{ntoinfty} frac{n-1}{n} = lim_{ntoinfty} 1 - frac 1 n = lim_{ntoinfty} 1 - lim_{ntoinfty} frac 1 n = 1 - 0 = 1$$



      Thus we can conclude that the continued fraction equals $1$, because its sequence of convergents equals $1$!




      On Infinite Radicals:



      So now, we touch on infinite nested radicals. They're messier to deal with but doable.



      One of the simpler examples of such radicals to contend with is



      $$2 = sqrt{2 +sqrt{2 +sqrt{2 +sqrt{2 +sqrt{2 +...}}}}}$$



      As with the previous two cases we see an infinite expression. We instinctively conclude by now: to logically define a limit for this expression - to assign it a value provided it even exists - we need to chop this up at finite points, defining a sequence of convergents $C(n)$, and then find $C(n)$ as $ntoinfty$.



      Nested radicals are a lot messier than the previous, but we manage.



      So first let the sequence of convergents be given by cutting off everything after the $n^{th} 2$ in the expression. Thus we get the sequence



      $$sqrt 2 ;;,;; sqrt{2 + sqrt{2}};;,;; sqrt{2+sqrt{2+sqrt{2}}};;,;; sqrt{2+sqrt{2+sqrt{2+sqrt{2}}}}$$



      Okay this isn't particularly nice already, but apparently there does exist, shockingly enough, a closed-form explicit expression for $C(n)$: (from: S. Zimmerman, C. Ho)



      $$C(n) = 2cosleft(frac{pi}{2^{n+1}}right)$$



      (I had to find that expression by Googling, I honestly didn't know that offhand. It can be proved by induction, as touched on in this MSE question.)



      So luckily, then, we can find the limit of $C(n)$:



      $$lim_{ntoinfty} C(n) = lim_{ntoinfty} 2cosleft(frac{pi}{2^{n+1}}right)$$



      It is probably obvious that the argument of the cosine function approaches $0$ as $n$ grows without bound, and thus



      $$lim_{ntoinfty} C(n) = lim_{ntoinfty} 2cosleft(frac{pi}{2^{n+1}}right) = lim_{ntoinfty} 2cos(0) = 2cdot 1 = 2$$



      Thus, since its convergents approach $2$, we can conclude that



      $$2 = sqrt{2 +sqrt{2 +sqrt{2 +sqrt{2 +sqrt{2 +...}}}}}$$




      A Lengthy Conclusion:



      So, in short, how do we evaluation an infinite expression, be it radical, continued fraction, sum, or otherwise?



      We begin by truncating the expression at convenient finite places, creating a series of convergents, generalizations of the "partial sums" introduced in calculus. We then try to get a closed form or some other usable expression for the convergents $C(n)$, and consider the value as $ntoinfty$. If it converges to some value, we say that the expression is in fact equal to that value. If it doesn't, then the expression doesn't converge to any value.



      This doesn't mean each expression is "nice." Radical expressions in particular, in my experience, tend to be nasty as all hell, and I'm lucky I found that one closed form expression for the particularly easy radical I chose.



      This doesn't mean that other methods cannot be used to find the values, so long as there's some sort of logical justification for said method. For example, there is a justification for the formula for an infinite (and finite) geometric sum. We might have to circumvent the notion of partial sums entirely, or at least it might be convenient to do so. For example, with the Basel problem, Euler's proof focused on Maclaurin series, and none of this "convergent" stuff. (That proof is here plus other proofs of it!)



      Luckily, at least, this notion of convergents, even if it may not always be the best way to do it, lends itself to an easy way to check a solution to any such problem. Just find a bunch of the convergents - take as many as you need. If you somehow have multiple solutions, as you think with Ramanujan's radical, then you'll see the convergents get closer and closer to the "true" solution. (How many convergents you need to find depends on the situation and how close your proposed solutions are. It might be immediately obvious after $10$ iterations, or might not be until $10,000,000$.)



      So I will conclude by showing, in this way, that the solution is $3$ to Ramanujan's radical.



      We begin with the radical itself:



      $$sqrt{1+2sqrt{1+3sqrt{1+4sqrt{1+cdots}}}}=3$$



      Let us begin by getting a series of convergents:



      $$sqrt{1} ;;,;; sqrt{1 + 2sqrt{1}} ;;,;; sqrt{1 + 2sqrt{1 + 3sqrt{1}}} ;;,;;$$



      Because the $sqrt{1}$ isn't necessary, we just let it be $1$.



      $$1 ;;,;; sqrt{1 + 2} ;;,;; sqrt{1 + 2sqrt{1 + 3}} ;;,;;$$



      Okay so ... where to go from here? Honestly, my initial temptation was to just use a MATLAB script and evaluate it, but I can't think of even a recursive closed form for this that would be nice enough. So in any event, we just have to go by "hand" (and by hand I mean WolframAlpha). Let $C(n)$ be the $n^{th}$ convergent. Then




      • $C(1) = 1$

      • $C(2) approx 1.732$

      • $C(3) approx 2.236$

      • $C(4) approx 2.560$

      • $C(5) approx 2.755$

      • $C(6) approx 2.867$

      • $C(7) approx 2.929$

      • $C(8) approx 2.963$

      • $C(9) approx 2.981$

      • $C(10) approx 2.990$


      To skip a few values because at this point the changes get minimal, I used a macro to make a quick code for $C(50)$ so I could put it into Microsoft Excel and got the approximate result



      $$C(50) approx 2.999 ; 999 ; 999 ; 999 ; 99$$



      So while not the most rigorous result, we can at least on an intuitive level feel like the convergents from Ramanujan's radical converge to $3$, not $4$ or any other number. Neglecting that this is not an ironclad proof of the convergence, at least intuitively then we can feel like



      $$3 = sqrt{1+2sqrt{1+3sqrt{1+4sqrt{1+cdots}}}}$$




      Whew! Hopefully that lengthy post was helpful to you!




      A late footnote, but Mathologer on YouTube did a video on this very topic, so his video would give a decent summary of all this stuff as well. Here's a link.






      share|cite|improve this answer











      $endgroup$


















        3












        $begingroup$

        Introduction:



        The issue is what "..." really "represents."



        Typically we use it as a sort of shorthand, as if to say "look, I can't write infinitely many things down, just assume that the obvious pattern holds and goes on infinitely."



        This idea holds for all sorts of things - nested radicals, infinite sums, continued fractions, infinite sequences, etc.




        On Infinite Sums:



        A simple example: the sum of the reciprocals of squares:



        $$1 + frac{1}{4} + frac{1}{9} + frac{1}{16} + ...$$



        This is a well known summation. It is the Riemann zeta function $zeta(s)$ at $s=2$, and is known to evaluate to $pi^2/6$ (proved by Euler and known as the Basel problem).



        Another, easier-to-handle summation is the geometric sum



        $$1 + frac 1 2 + frac 1 4 + frac 1 8 + ...$$



        This is a geometric series where the ratio is $1/2$ - each summand is half the previous one. We know, too, that this evaluates to $2$.



        Another geometric series you might see in proofs that $0.999... = 1$ is



        $$9 left( frac{1}{10} + frac{1}{100} + frac{1}{1,000} + frac{1}{10,000} + ... right)$$



        which equals $1$. In fact, any infinite geometric series, with first term $a$ and ratio $|r|<1$ can be evaluated by



        $$sum_{n=0}^infty ar^n = frac{a}{1-r}$$



        So a question arises - ignoring these "obvious" results (depending on your amount of mathematical knowledge), how would we know these converge to the given values? What, exactly, does it mean for a summation to converge to a number or equal a number? For finite sums this is no issue - if nothing else, we could add up each number manually, but we can't just add up every number from a set of infinitely-many numbers.



        Well, one could argue by common sense that, if the sequence converges to some number, the more and more terms you add up, the closer they'll get to that number.



        So we obtain one definition for the convergence of an infinite sum. Consider a sequence where the $n^{th}$ term is defined by the sum of the first $n$ terms in the sequence. To introduce some symbols, suppose we're trying to find the sum



        $$sum_{k=1}^infty x_k = x_1 + x_2 + x_3 + x_4 + ...$$



        for whatever these $x_i$'s are. Then define these so-called "partial sums" of this by a function $S(n)$:



        $$S(n) = sum_{k=1}^n x_k = x_1 + x_2 + ... + x_n$$



        Then we get a sequence of sums:



        $$S_1, S_2, S_3, ...$$



        or equivalently



        $$x_1 ;;,;; x_1 + x_2;;,;; x_1 + x_2 + x_3;;,;; ...$$



        Then we ask: what does $S(n)$ approach as $n$ grows without bound, if anything at all? (In calculus, we call this "the limit of the partial sums $S(n)$ as $n$ approaches infinity.")



        For the case of our first geometric sum, we immediately see the sequence of partial sums



        $$1, frac{3}{2}, frac{5}{4}, frac{15}{8},...$$



        Clearly, this suggests a pattern - and if you want to, you can go ahead and prove it, I won't do so here for brevity's sake. The pattern is that the $n^{th}$ term of the sequence is



        $$S(n) = frac{2n-1}{n}$$



        We can then easily consider the limit of these partial sums:



        $$lim_{ntoinfty} S(n) = lim_{ntoinfty} frac{2n-1}{n} = lim_{ntoinfty} 2 - frac 1 n = lim_{ntoinfty} 2 - lim_{ntoinfty} frac 1 n$$



        Obviously, $1/n to 0$ as $n$ grows without bound, and $2$ is not affected by $n$, so we conclude



        $$lim_{ntoinfty} S(n) = lim_{ntoinfty} 2 - lim_{ntoinfty} frac 1 n = 2 - 0 = 2$$



        And thus we say



        $$sum_{k=0}^infty left(frac 1 2 right)^k = 1 + frac 1 2 + frac 1 4 + frac 1 8 + ... = 2$$



        because the partial sums approach $2$.




        On Continued Fractions:



        That was a simple, "first" sort of example, but mathematicians essentially do the same thing in other contexts. I want to touch on one more such context before we deal with the radical case, just to nail that point home.



        In this case, it will be with continued fractions. One of the simpler such fractions is the one for $1$:



        $$1 = frac{1}{2-frac{1}{2-frac{1}{2-frac{1}{...}}}}$$



        As usual, the "..." denotes that this continues forever. But what it does it mean for this infinite expression to equal $1$?



        For this, we consider a more general analogue of the "partial sum" from before - a "convergent." We cut up the sequence at logical finite points, whatever those points being depending on the context. Then if the sequence of the convergents approaches a limit, we say they're equal.



        What are the convergents for a continued fraction? By convention, we cut off just before the start of the next fraction. That is, in the continued fraction for $1$, we cut off at the $n^{th} 2$ for the $n^{th}$ convergent, and ignore what follows. So we get the sequence of convergents



        $$frac{1}{2} , frac{1}{2-frac{1}{2}}, frac{1}{2-frac{1}{2-frac{1}{2}}}$$



        Working out the numbers, we find the sequence to be



        $$frac{1}{2},frac{2}{3},frac{3}{4},...$$



        Again, we see a pattern! The $n^{th}$ term of the sequence is clearly of the form



        $$frac{n-1}{n}$$



        Let $C(n)$ be a function denoting the $n^{th}$ convergent. Then $C(1)=1/2,$ $C(2) = 2/3,$ $C(n)=(n-1)/n,$ and so on. So like before we consider the infinite limit:



        $$lim_{ntoinfty} C(n) = lim_{ntoinfty} frac{n-1}{n} = lim_{ntoinfty} 1 - frac 1 n = lim_{ntoinfty} 1 - lim_{ntoinfty} frac 1 n = 1 - 0 = 1$$



        Thus we can conclude that the continued fraction equals $1$, because its sequence of convergents equals $1$!




        On Infinite Radicals:



        So now, we touch on infinite nested radicals. They're messier to deal with but doable.



        One of the simpler examples of such radicals to contend with is



        $$2 = sqrt{2 +sqrt{2 +sqrt{2 +sqrt{2 +sqrt{2 +...}}}}}$$



        As with the previous two cases we see an infinite expression. We instinctively conclude by now: to logically define a limit for this expression - to assign it a value provided it even exists - we need to chop this up at finite points, defining a sequence of convergents $C(n)$, and then find $C(n)$ as $ntoinfty$.



        Nested radicals are a lot messier than the previous, but we manage.



        So first let the sequence of convergents be given by cutting off everything after the $n^{th} 2$ in the expression. Thus we get the sequence



        $$sqrt 2 ;;,;; sqrt{2 + sqrt{2}};;,;; sqrt{2+sqrt{2+sqrt{2}}};;,;; sqrt{2+sqrt{2+sqrt{2+sqrt{2}}}}$$



        Okay this isn't particularly nice already, but apparently there does exist, shockingly enough, a closed-form explicit expression for $C(n)$: (from: S. Zimmerman, C. Ho)



        $$C(n) = 2cosleft(frac{pi}{2^{n+1}}right)$$



        (I had to find that expression by Googling, I honestly didn't know that offhand. It can be proved by induction, as touched on in this MSE question.)



        So luckily, then, we can find the limit of $C(n)$:



        $$lim_{ntoinfty} C(n) = lim_{ntoinfty} 2cosleft(frac{pi}{2^{n+1}}right)$$



        It is probably obvious that the argument of the cosine function approaches $0$ as $n$ grows without bound, and thus



        $$lim_{ntoinfty} C(n) = lim_{ntoinfty} 2cosleft(frac{pi}{2^{n+1}}right) = lim_{ntoinfty} 2cos(0) = 2cdot 1 = 2$$



        Thus, since its convergents approach $2$, we can conclude that



        $$2 = sqrt{2 +sqrt{2 +sqrt{2 +sqrt{2 +sqrt{2 +...}}}}}$$




        A Lengthy Conclusion:



        So, in short, how do we evaluation an infinite expression, be it radical, continued fraction, sum, or otherwise?



        We begin by truncating the expression at convenient finite places, creating a series of convergents, generalizations of the "partial sums" introduced in calculus. We then try to get a closed form or some other usable expression for the convergents $C(n)$, and consider the value as $ntoinfty$. If it converges to some value, we say that the expression is in fact equal to that value. If it doesn't, then the expression doesn't converge to any value.



        This doesn't mean each expression is "nice." Radical expressions in particular, in my experience, tend to be nasty as all hell, and I'm lucky I found that one closed form expression for the particularly easy radical I chose.



        This doesn't mean that other methods cannot be used to find the values, so long as there's some sort of logical justification for said method. For example, there is a justification for the formula for an infinite (and finite) geometric sum. We might have to circumvent the notion of partial sums entirely, or at least it might be convenient to do so. For example, with the Basel problem, Euler's proof focused on Maclaurin series, and none of this "convergent" stuff. (That proof is here plus other proofs of it!)



        Luckily, at least, this notion of convergents, even if it may not always be the best way to do it, lends itself to an easy way to check a solution to any such problem. Just find a bunch of the convergents - take as many as you need. If you somehow have multiple solutions, as you think with Ramanujan's radical, then you'll see the convergents get closer and closer to the "true" solution. (How many convergents you need to find depends on the situation and how close your proposed solutions are. It might be immediately obvious after $10$ iterations, or might not be until $10,000,000$.)



        So I will conclude by showing, in this way, that the solution is $3$ to Ramanujan's radical.



        We begin with the radical itself:



        $$sqrt{1+2sqrt{1+3sqrt{1+4sqrt{1+cdots}}}}=3$$



        Let us begin by getting a series of convergents:



        $$sqrt{1} ;;,;; sqrt{1 + 2sqrt{1}} ;;,;; sqrt{1 + 2sqrt{1 + 3sqrt{1}}} ;;,;;$$



        Because the $sqrt{1}$ isn't necessary, we just let it be $1$.



        $$1 ;;,;; sqrt{1 + 2} ;;,;; sqrt{1 + 2sqrt{1 + 3}} ;;,;;$$



        Okay so ... where to go from here? Honestly, my initial temptation was to just use a MATLAB script and evaluate it, but I can't think of even a recursive closed form for this that would be nice enough. So in any event, we just have to go by "hand" (and by hand I mean WolframAlpha). Let $C(n)$ be the $n^{th}$ convergent. Then




        • $C(1) = 1$

        • $C(2) approx 1.732$

        • $C(3) approx 2.236$

        • $C(4) approx 2.560$

        • $C(5) approx 2.755$

        • $C(6) approx 2.867$

        • $C(7) approx 2.929$

        • $C(8) approx 2.963$

        • $C(9) approx 2.981$

        • $C(10) approx 2.990$


        To skip a few values because at this point the changes get minimal, I used a macro to make a quick code for $C(50)$ so I could put it into Microsoft Excel and got the approximate result



        $$C(50) approx 2.999 ; 999 ; 999 ; 999 ; 99$$



        So while not the most rigorous result, we can at least on an intuitive level feel like the convergents from Ramanujan's radical converge to $3$, not $4$ or any other number. Neglecting that this is not an ironclad proof of the convergence, at least intuitively then we can feel like



        $$3 = sqrt{1+2sqrt{1+3sqrt{1+4sqrt{1+cdots}}}}$$




        Whew! Hopefully that lengthy post was helpful to you!




        A late footnote, but Mathologer on YouTube did a video on this very topic, so his video would give a decent summary of all this stuff as well. Here's a link.






        share|cite|improve this answer











        $endgroup$
















          3












          3








          3





          $begingroup$

          Introduction:



          The issue is what "..." really "represents."



          Typically we use it as a sort of shorthand, as if to say "look, I can't write infinitely many things down, just assume that the obvious pattern holds and goes on infinitely."



          This idea holds for all sorts of things - nested radicals, infinite sums, continued fractions, infinite sequences, etc.




          On Infinite Sums:



          A simple example: the sum of the reciprocals of squares:



          $$1 + frac{1}{4} + frac{1}{9} + frac{1}{16} + ...$$



          This is a well known summation. It is the Riemann zeta function $zeta(s)$ at $s=2$, and is known to evaluate to $pi^2/6$ (proved by Euler and known as the Basel problem).



          Another, easier-to-handle summation is the geometric sum



          $$1 + frac 1 2 + frac 1 4 + frac 1 8 + ...$$



          This is a geometric series where the ratio is $1/2$ - each summand is half the previous one. We know, too, that this evaluates to $2$.



          Another geometric series you might see in proofs that $0.999... = 1$ is



          $$9 left( frac{1}{10} + frac{1}{100} + frac{1}{1,000} + frac{1}{10,000} + ... right)$$



          which equals $1$. In fact, any infinite geometric series, with first term $a$ and ratio $|r|<1$ can be evaluated by



          $$sum_{n=0}^infty ar^n = frac{a}{1-r}$$



          So a question arises - ignoring these "obvious" results (depending on your amount of mathematical knowledge), how would we know these converge to the given values? What, exactly, does it mean for a summation to converge to a number or equal a number? For finite sums this is no issue - if nothing else, we could add up each number manually, but we can't just add up every number from a set of infinitely-many numbers.



          Well, one could argue by common sense that, if the sequence converges to some number, the more and more terms you add up, the closer they'll get to that number.



          So we obtain one definition for the convergence of an infinite sum. Consider a sequence where the $n^{th}$ term is defined by the sum of the first $n$ terms in the sequence. To introduce some symbols, suppose we're trying to find the sum



          $$sum_{k=1}^infty x_k = x_1 + x_2 + x_3 + x_4 + ...$$



          for whatever these $x_i$'s are. Then define these so-called "partial sums" of this by a function $S(n)$:



          $$S(n) = sum_{k=1}^n x_k = x_1 + x_2 + ... + x_n$$



          Then we get a sequence of sums:



          $$S_1, S_2, S_3, ...$$



          or equivalently



          $$x_1 ;;,;; x_1 + x_2;;,;; x_1 + x_2 + x_3;;,;; ...$$



          Then we ask: what does $S(n)$ approach as $n$ grows without bound, if anything at all? (In calculus, we call this "the limit of the partial sums $S(n)$ as $n$ approaches infinity.")



          For the case of our first geometric sum, we immediately see the sequence of partial sums



          $$1, frac{3}{2}, frac{5}{4}, frac{15}{8},...$$



          Clearly, this suggests a pattern - and if you want to, you can go ahead and prove it, I won't do so here for brevity's sake. The pattern is that the $n^{th}$ term of the sequence is



          $$S(n) = frac{2n-1}{n}$$



          We can then easily consider the limit of these partial sums:



          $$lim_{ntoinfty} S(n) = lim_{ntoinfty} frac{2n-1}{n} = lim_{ntoinfty} 2 - frac 1 n = lim_{ntoinfty} 2 - lim_{ntoinfty} frac 1 n$$



          Obviously, $1/n to 0$ as $n$ grows without bound, and $2$ is not affected by $n$, so we conclude



          $$lim_{ntoinfty} S(n) = lim_{ntoinfty} 2 - lim_{ntoinfty} frac 1 n = 2 - 0 = 2$$



          And thus we say



          $$sum_{k=0}^infty left(frac 1 2 right)^k = 1 + frac 1 2 + frac 1 4 + frac 1 8 + ... = 2$$



          because the partial sums approach $2$.




          On Continued Fractions:



          That was a simple, "first" sort of example, but mathematicians essentially do the same thing in other contexts. I want to touch on one more such context before we deal with the radical case, just to nail that point home.



          In this case, it will be with continued fractions. One of the simpler such fractions is the one for $1$:



          $$1 = frac{1}{2-frac{1}{2-frac{1}{2-frac{1}{...}}}}$$



          As usual, the "..." denotes that this continues forever. But what it does it mean for this infinite expression to equal $1$?



          For this, we consider a more general analogue of the "partial sum" from before - a "convergent." We cut up the sequence at logical finite points, whatever those points being depending on the context. Then if the sequence of the convergents approaches a limit, we say they're equal.



          What are the convergents for a continued fraction? By convention, we cut off just before the start of the next fraction. That is, in the continued fraction for $1$, we cut off at the $n^{th} 2$ for the $n^{th}$ convergent, and ignore what follows. So we get the sequence of convergents



          $$frac{1}{2} , frac{1}{2-frac{1}{2}}, frac{1}{2-frac{1}{2-frac{1}{2}}}$$



          Working out the numbers, we find the sequence to be



          $$frac{1}{2},frac{2}{3},frac{3}{4},...$$



          Again, we see a pattern! The $n^{th}$ term of the sequence is clearly of the form



          $$frac{n-1}{n}$$



          Let $C(n)$ be a function denoting the $n^{th}$ convergent. Then $C(1)=1/2,$ $C(2) = 2/3,$ $C(n)=(n-1)/n,$ and so on. So like before we consider the infinite limit:



          $$lim_{ntoinfty} C(n) = lim_{ntoinfty} frac{n-1}{n} = lim_{ntoinfty} 1 - frac 1 n = lim_{ntoinfty} 1 - lim_{ntoinfty} frac 1 n = 1 - 0 = 1$$



          Thus we can conclude that the continued fraction equals $1$, because its sequence of convergents equals $1$!




          On Infinite Radicals:



          So now, we touch on infinite nested radicals. They're messier to deal with but doable.



          One of the simpler examples of such radicals to contend with is



          $$2 = sqrt{2 +sqrt{2 +sqrt{2 +sqrt{2 +sqrt{2 +...}}}}}$$



          As with the previous two cases we see an infinite expression. We instinctively conclude by now: to logically define a limit for this expression - to assign it a value provided it even exists - we need to chop this up at finite points, defining a sequence of convergents $C(n)$, and then find $C(n)$ as $ntoinfty$.



          Nested radicals are a lot messier than the previous, but we manage.



          So first let the sequence of convergents be given by cutting off everything after the $n^{th} 2$ in the expression. Thus we get the sequence



          $$sqrt 2 ;;,;; sqrt{2 + sqrt{2}};;,;; sqrt{2+sqrt{2+sqrt{2}}};;,;; sqrt{2+sqrt{2+sqrt{2+sqrt{2}}}}$$



          Okay this isn't particularly nice already, but apparently there does exist, shockingly enough, a closed-form explicit expression for $C(n)$: (from: S. Zimmerman, C. Ho)



          $$C(n) = 2cosleft(frac{pi}{2^{n+1}}right)$$



          (I had to find that expression by Googling, I honestly didn't know that offhand. It can be proved by induction, as touched on in this MSE question.)



          So luckily, then, we can find the limit of $C(n)$:



          $$lim_{ntoinfty} C(n) = lim_{ntoinfty} 2cosleft(frac{pi}{2^{n+1}}right)$$



          It is probably obvious that the argument of the cosine function approaches $0$ as $n$ grows without bound, and thus



          $$lim_{ntoinfty} C(n) = lim_{ntoinfty} 2cosleft(frac{pi}{2^{n+1}}right) = lim_{ntoinfty} 2cos(0) = 2cdot 1 = 2$$



          Thus, since its convergents approach $2$, we can conclude that



          $$2 = sqrt{2 +sqrt{2 +sqrt{2 +sqrt{2 +sqrt{2 +...}}}}}$$




          A Lengthy Conclusion:



          So, in short, how do we evaluation an infinite expression, be it radical, continued fraction, sum, or otherwise?



          We begin by truncating the expression at convenient finite places, creating a series of convergents, generalizations of the "partial sums" introduced in calculus. We then try to get a closed form or some other usable expression for the convergents $C(n)$, and consider the value as $ntoinfty$. If it converges to some value, we say that the expression is in fact equal to that value. If it doesn't, then the expression doesn't converge to any value.



          This doesn't mean each expression is "nice." Radical expressions in particular, in my experience, tend to be nasty as all hell, and I'm lucky I found that one closed form expression for the particularly easy radical I chose.



          This doesn't mean that other methods cannot be used to find the values, so long as there's some sort of logical justification for said method. For example, there is a justification for the formula for an infinite (and finite) geometric sum. We might have to circumvent the notion of partial sums entirely, or at least it might be convenient to do so. For example, with the Basel problem, Euler's proof focused on Maclaurin series, and none of this "convergent" stuff. (That proof is here plus other proofs of it!)



          Luckily, at least, this notion of convergents, even if it may not always be the best way to do it, lends itself to an easy way to check a solution to any such problem. Just find a bunch of the convergents - take as many as you need. If you somehow have multiple solutions, as you think with Ramanujan's radical, then you'll see the convergents get closer and closer to the "true" solution. (How many convergents you need to find depends on the situation and how close your proposed solutions are. It might be immediately obvious after $10$ iterations, or might not be until $10,000,000$.)



          So I will conclude by showing, in this way, that the solution is $3$ to Ramanujan's radical.



          We begin with the radical itself:



          $$sqrt{1+2sqrt{1+3sqrt{1+4sqrt{1+cdots}}}}=3$$



          Let us begin by getting a series of convergents:



          $$sqrt{1} ;;,;; sqrt{1 + 2sqrt{1}} ;;,;; sqrt{1 + 2sqrt{1 + 3sqrt{1}}} ;;,;;$$



          Because the $sqrt{1}$ isn't necessary, we just let it be $1$.



          $$1 ;;,;; sqrt{1 + 2} ;;,;; sqrt{1 + 2sqrt{1 + 3}} ;;,;;$$



          Okay so ... where to go from here? Honestly, my initial temptation was to just use a MATLAB script and evaluate it, but I can't think of even a recursive closed form for this that would be nice enough. So in any event, we just have to go by "hand" (and by hand I mean WolframAlpha). Let $C(n)$ be the $n^{th}$ convergent. Then




          • $C(1) = 1$

          • $C(2) approx 1.732$

          • $C(3) approx 2.236$

          • $C(4) approx 2.560$

          • $C(5) approx 2.755$

          • $C(6) approx 2.867$

          • $C(7) approx 2.929$

          • $C(8) approx 2.963$

          • $C(9) approx 2.981$

          • $C(10) approx 2.990$


          To skip a few values because at this point the changes get minimal, I used a macro to make a quick code for $C(50)$ so I could put it into Microsoft Excel and got the approximate result



          $$C(50) approx 2.999 ; 999 ; 999 ; 999 ; 99$$



          So while not the most rigorous result, we can at least on an intuitive level feel like the convergents from Ramanujan's radical converge to $3$, not $4$ or any other number. Neglecting that this is not an ironclad proof of the convergence, at least intuitively then we can feel like



          $$3 = sqrt{1+2sqrt{1+3sqrt{1+4sqrt{1+cdots}}}}$$




          Whew! Hopefully that lengthy post was helpful to you!




          A late footnote, but Mathologer on YouTube did a video on this very topic, so his video would give a decent summary of all this stuff as well. Here's a link.






          share|cite|improve this answer











          $endgroup$



          Introduction:



          The issue is what "..." really "represents."



          Typically we use it as a sort of shorthand, as if to say "look, I can't write infinitely many things down, just assume that the obvious pattern holds and goes on infinitely."



          This idea holds for all sorts of things - nested radicals, infinite sums, continued fractions, infinite sequences, etc.




          On Infinite Sums:



          A simple example: the sum of the reciprocals of squares:



          $$1 + frac{1}{4} + frac{1}{9} + frac{1}{16} + ...$$



          This is a well known summation. It is the Riemann zeta function $zeta(s)$ at $s=2$, and is known to evaluate to $pi^2/6$ (proved by Euler and known as the Basel problem).



          Another, easier-to-handle summation is the geometric sum



          $$1 + frac 1 2 + frac 1 4 + frac 1 8 + ...$$



          This is a geometric series where the ratio is $1/2$ - each summand is half the previous one. We know, too, that this evaluates to $2$.



          Another geometric series you might see in proofs that $0.999... = 1$ is



          $$9 left( frac{1}{10} + frac{1}{100} + frac{1}{1,000} + frac{1}{10,000} + ... right)$$



          which equals $1$. In fact, any infinite geometric series, with first term $a$ and ratio $|r|<1$ can be evaluated by



          $$sum_{n=0}^infty ar^n = frac{a}{1-r}$$



          So a question arises - ignoring these "obvious" results (depending on your amount of mathematical knowledge), how would we know these converge to the given values? What, exactly, does it mean for a summation to converge to a number or equal a number? For finite sums this is no issue - if nothing else, we could add up each number manually, but we can't just add up every number from a set of infinitely-many numbers.



          Well, one could argue by common sense that, if the sequence converges to some number, the more and more terms you add up, the closer they'll get to that number.



          So we obtain one definition for the convergence of an infinite sum. Consider a sequence where the $n^{th}$ term is defined by the sum of the first $n$ terms in the sequence. To introduce some symbols, suppose we're trying to find the sum



          $$sum_{k=1}^infty x_k = x_1 + x_2 + x_3 + x_4 + ...$$



          for whatever these $x_i$'s are. Then define these so-called "partial sums" of this by a function $S(n)$:



          $$S(n) = sum_{k=1}^n x_k = x_1 + x_2 + ... + x_n$$



          Then we get a sequence of sums:



          $$S_1, S_2, S_3, ...$$



          or equivalently



          $$x_1 ;;,;; x_1 + x_2;;,;; x_1 + x_2 + x_3;;,;; ...$$



          Then we ask: what does $S(n)$ approach as $n$ grows without bound, if anything at all? (In calculus, we call this "the limit of the partial sums $S(n)$ as $n$ approaches infinity.")



          For the case of our first geometric sum, we immediately see the sequence of partial sums



          $$1, frac{3}{2}, frac{5}{4}, frac{15}{8},...$$



          Clearly, this suggests a pattern - and if you want to, you can go ahead and prove it, I won't do so here for brevity's sake. The pattern is that the $n^{th}$ term of the sequence is



          $$S(n) = frac{2n-1}{n}$$



          We can then easily consider the limit of these partial sums:



          $$lim_{ntoinfty} S(n) = lim_{ntoinfty} frac{2n-1}{n} = lim_{ntoinfty} 2 - frac 1 n = lim_{ntoinfty} 2 - lim_{ntoinfty} frac 1 n$$



          Obviously, $1/n to 0$ as $n$ grows without bound, and $2$ is not affected by $n$, so we conclude



          $$lim_{ntoinfty} S(n) = lim_{ntoinfty} 2 - lim_{ntoinfty} frac 1 n = 2 - 0 = 2$$



          And thus we say



          $$sum_{k=0}^infty left(frac 1 2 right)^k = 1 + frac 1 2 + frac 1 4 + frac 1 8 + ... = 2$$



          because the partial sums approach $2$.




          On Continued Fractions:



          That was a simple, "first" sort of example, but mathematicians essentially do the same thing in other contexts. I want to touch on one more such context before we deal with the radical case, just to nail that point home.



          In this case, it will be with continued fractions. One of the simpler such fractions is the one for $1$:



          $$1 = frac{1}{2-frac{1}{2-frac{1}{2-frac{1}{...}}}}$$



          As usual, the "..." denotes that this continues forever. But what it does it mean for this infinite expression to equal $1$?



          For this, we consider a more general analogue of the "partial sum" from before - a "convergent." We cut up the sequence at logical finite points, whatever those points being depending on the context. Then if the sequence of the convergents approaches a limit, we say they're equal.



          What are the convergents for a continued fraction? By convention, we cut off just before the start of the next fraction. That is, in the continued fraction for $1$, we cut off at the $n^{th} 2$ for the $n^{th}$ convergent, and ignore what follows. So we get the sequence of convergents



          $$frac{1}{2} , frac{1}{2-frac{1}{2}}, frac{1}{2-frac{1}{2-frac{1}{2}}}$$



          Working out the numbers, we find the sequence to be



          $$frac{1}{2},frac{2}{3},frac{3}{4},...$$



          Again, we see a pattern! The $n^{th}$ term of the sequence is clearly of the form



          $$frac{n-1}{n}$$



          Let $C(n)$ be a function denoting the $n^{th}$ convergent. Then $C(1)=1/2,$ $C(2) = 2/3,$ $C(n)=(n-1)/n,$ and so on. So like before we consider the infinite limit:



          $$lim_{ntoinfty} C(n) = lim_{ntoinfty} frac{n-1}{n} = lim_{ntoinfty} 1 - frac 1 n = lim_{ntoinfty} 1 - lim_{ntoinfty} frac 1 n = 1 - 0 = 1$$



          Thus we can conclude that the continued fraction equals $1$, because its sequence of convergents equals $1$!




          On Infinite Radicals:



          So now, we touch on infinite nested radicals. They're messier to deal with but doable.



          One of the simpler examples of such radicals to contend with is



          $$2 = sqrt{2 +sqrt{2 +sqrt{2 +sqrt{2 +sqrt{2 +...}}}}}$$



          As with the previous two cases we see an infinite expression. We instinctively conclude by now: to logically define a limit for this expression - to assign it a value provided it even exists - we need to chop this up at finite points, defining a sequence of convergents $C(n)$, and then find $C(n)$ as $ntoinfty$.



          Nested radicals are a lot messier than the previous, but we manage.



          So first let the sequence of convergents be given by cutting off everything after the $n^{th} 2$ in the expression. Thus we get the sequence



          $$sqrt 2 ;;,;; sqrt{2 + sqrt{2}};;,;; sqrt{2+sqrt{2+sqrt{2}}};;,;; sqrt{2+sqrt{2+sqrt{2+sqrt{2}}}}$$



          Okay this isn't particularly nice already, but apparently there does exist, shockingly enough, a closed-form explicit expression for $C(n)$: (from: S. Zimmerman, C. Ho)



          $$C(n) = 2cosleft(frac{pi}{2^{n+1}}right)$$



          (I had to find that expression by Googling, I honestly didn't know that offhand. It can be proved by induction, as touched on in this MSE question.)



          So luckily, then, we can find the limit of $C(n)$:



          $$lim_{ntoinfty} C(n) = lim_{ntoinfty} 2cosleft(frac{pi}{2^{n+1}}right)$$



          It is probably obvious that the argument of the cosine function approaches $0$ as $n$ grows without bound, and thus



          $$lim_{ntoinfty} C(n) = lim_{ntoinfty} 2cosleft(frac{pi}{2^{n+1}}right) = lim_{ntoinfty} 2cos(0) = 2cdot 1 = 2$$



          Thus, since its convergents approach $2$, we can conclude that



          $$2 = sqrt{2 +sqrt{2 +sqrt{2 +sqrt{2 +sqrt{2 +...}}}}}$$




          A Lengthy Conclusion:



          So, in short, how do we evaluation an infinite expression, be it radical, continued fraction, sum, or otherwise?



          We begin by truncating the expression at convenient finite places, creating a series of convergents, generalizations of the "partial sums" introduced in calculus. We then try to get a closed form or some other usable expression for the convergents $C(n)$, and consider the value as $ntoinfty$. If it converges to some value, we say that the expression is in fact equal to that value. If it doesn't, then the expression doesn't converge to any value.



          This doesn't mean each expression is "nice." Radical expressions in particular, in my experience, tend to be nasty as all hell, and I'm lucky I found that one closed form expression for the particularly easy radical I chose.



          This doesn't mean that other methods cannot be used to find the values, so long as there's some sort of logical justification for said method. For example, there is a justification for the formula for an infinite (and finite) geometric sum. We might have to circumvent the notion of partial sums entirely, or at least it might be convenient to do so. For example, with the Basel problem, Euler's proof focused on Maclaurin series, and none of this "convergent" stuff. (That proof is here plus other proofs of it!)



          Luckily, at least, this notion of convergents, even if it may not always be the best way to do it, lends itself to an easy way to check a solution to any such problem. Just find a bunch of the convergents - take as many as you need. If you somehow have multiple solutions, as you think with Ramanujan's radical, then you'll see the convergents get closer and closer to the "true" solution. (How many convergents you need to find depends on the situation and how close your proposed solutions are. It might be immediately obvious after $10$ iterations, or might not be until $10,000,000$.)



          So I will conclude by showing, in this way, that the solution is $3$ to Ramanujan's radical.



          We begin with the radical itself:



          $$sqrt{1+2sqrt{1+3sqrt{1+4sqrt{1+cdots}}}}=3$$



          Let us begin by getting a series of convergents:



          $$sqrt{1} ;;,;; sqrt{1 + 2sqrt{1}} ;;,;; sqrt{1 + 2sqrt{1 + 3sqrt{1}}} ;;,;;$$



          Because the $sqrt{1}$ isn't necessary, we just let it be $1$.



          $$1 ;;,;; sqrt{1 + 2} ;;,;; sqrt{1 + 2sqrt{1 + 3}} ;;,;;$$



          Okay so ... where to go from here? Honestly, my initial temptation was to just use a MATLAB script and evaluate it, but I can't think of even a recursive closed form for this that would be nice enough. So in any event, we just have to go by "hand" (and by hand I mean WolframAlpha). Let $C(n)$ be the $n^{th}$ convergent. Then




          • $C(1) = 1$

          • $C(2) approx 1.732$

          • $C(3) approx 2.236$

          • $C(4) approx 2.560$

          • $C(5) approx 2.755$

          • $C(6) approx 2.867$

          • $C(7) approx 2.929$

          • $C(8) approx 2.963$

          • $C(9) approx 2.981$

          • $C(10) approx 2.990$


          To skip a few values because at this point the changes get minimal, I used a macro to make a quick code for $C(50)$ so I could put it into Microsoft Excel and got the approximate result



          $$C(50) approx 2.999 ; 999 ; 999 ; 999 ; 99$$



          So while not the most rigorous result, we can at least on an intuitive level feel like the convergents from Ramanujan's radical converge to $3$, not $4$ or any other number. Neglecting that this is not an ironclad proof of the convergence, at least intuitively then we can feel like



          $$3 = sqrt{1+2sqrt{1+3sqrt{1+4sqrt{1+cdots}}}}$$




          Whew! Hopefully that lengthy post was helpful to you!




          A late footnote, but Mathologer on YouTube did a video on this very topic, so his video would give a decent summary of all this stuff as well. Here's a link.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 44 mins ago

























          answered 1 hour ago









          Eevee TrainerEevee Trainer

          6,1331936




          6,1331936























              0












              $begingroup$

              $4=sqrt{16}=sqrt{1+3sqrt{25}}=sqrt{1+3sqrt{1+4sqrt{36}}}=sqrt{1+3sqrt{1+4sqrt{1+5sqrt{49}}}}=sqrt{1+3sqrt{1+4sqrt{1+5sqrt{1+6sqrt{64}}}}}=sqrt{1+3sqrt{1+4sqrt{1+5sqrt{1+6sqrt{1+7sqrt{81}}}}}}$



              $=sqrt{1+3sqrt{1+4sqrt{1+5sqrt{1+6sqrt{1+7sqrt{1+8sqrt{1+cdots}}}}}}}$



              $5=sqrt{25}=sqrt{1+4sqrt{36}}=sqrt{1+4sqrt{1+5sqrt{49}}}=sqrt{1+4sqrt{1+5sqrt{1+6sqrt{64}}}}=sqrt{1+4sqrt{1+5sqrt{1+6sqrt{1+7sqrt{81}}}}}=sqrt{1+4sqrt{1+5sqrt{1+6sqrt{1+7sqrt{1+8sqrt{100}}}}}}$



              $=sqrt{1+4sqrt{1+5sqrt{1+6sqrt{1+7sqrt{1+8sqrt{1+9sqrt{1+cdots}}}}}}}$



              $vdots$



              $n=sqrt{1+(n-1)sqrt{1+nsqrt{1+(n+1)sqrt{1+(n+2)sqrt{1+(n+3)sqrt{1+(n+4)sqrt{1+cdots}}}}}}}$






              share|cite|improve this answer











              $endgroup$


















                0












                $begingroup$

                $4=sqrt{16}=sqrt{1+3sqrt{25}}=sqrt{1+3sqrt{1+4sqrt{36}}}=sqrt{1+3sqrt{1+4sqrt{1+5sqrt{49}}}}=sqrt{1+3sqrt{1+4sqrt{1+5sqrt{1+6sqrt{64}}}}}=sqrt{1+3sqrt{1+4sqrt{1+5sqrt{1+6sqrt{1+7sqrt{81}}}}}}$



                $=sqrt{1+3sqrt{1+4sqrt{1+5sqrt{1+6sqrt{1+7sqrt{1+8sqrt{1+cdots}}}}}}}$



                $5=sqrt{25}=sqrt{1+4sqrt{36}}=sqrt{1+4sqrt{1+5sqrt{49}}}=sqrt{1+4sqrt{1+5sqrt{1+6sqrt{64}}}}=sqrt{1+4sqrt{1+5sqrt{1+6sqrt{1+7sqrt{81}}}}}=sqrt{1+4sqrt{1+5sqrt{1+6sqrt{1+7sqrt{1+8sqrt{100}}}}}}$



                $=sqrt{1+4sqrt{1+5sqrt{1+6sqrt{1+7sqrt{1+8sqrt{1+9sqrt{1+cdots}}}}}}}$



                $vdots$



                $n=sqrt{1+(n-1)sqrt{1+nsqrt{1+(n+1)sqrt{1+(n+2)sqrt{1+(n+3)sqrt{1+(n+4)sqrt{1+cdots}}}}}}}$






                share|cite|improve this answer











                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  $4=sqrt{16}=sqrt{1+3sqrt{25}}=sqrt{1+3sqrt{1+4sqrt{36}}}=sqrt{1+3sqrt{1+4sqrt{1+5sqrt{49}}}}=sqrt{1+3sqrt{1+4sqrt{1+5sqrt{1+6sqrt{64}}}}}=sqrt{1+3sqrt{1+4sqrt{1+5sqrt{1+6sqrt{1+7sqrt{81}}}}}}$



                  $=sqrt{1+3sqrt{1+4sqrt{1+5sqrt{1+6sqrt{1+7sqrt{1+8sqrt{1+cdots}}}}}}}$



                  $5=sqrt{25}=sqrt{1+4sqrt{36}}=sqrt{1+4sqrt{1+5sqrt{49}}}=sqrt{1+4sqrt{1+5sqrt{1+6sqrt{64}}}}=sqrt{1+4sqrt{1+5sqrt{1+6sqrt{1+7sqrt{81}}}}}=sqrt{1+4sqrt{1+5sqrt{1+6sqrt{1+7sqrt{1+8sqrt{100}}}}}}$



                  $=sqrt{1+4sqrt{1+5sqrt{1+6sqrt{1+7sqrt{1+8sqrt{1+9sqrt{1+cdots}}}}}}}$



                  $vdots$



                  $n=sqrt{1+(n-1)sqrt{1+nsqrt{1+(n+1)sqrt{1+(n+2)sqrt{1+(n+3)sqrt{1+(n+4)sqrt{1+cdots}}}}}}}$






                  share|cite|improve this answer











                  $endgroup$



                  $4=sqrt{16}=sqrt{1+3sqrt{25}}=sqrt{1+3sqrt{1+4sqrt{36}}}=sqrt{1+3sqrt{1+4sqrt{1+5sqrt{49}}}}=sqrt{1+3sqrt{1+4sqrt{1+5sqrt{1+6sqrt{64}}}}}=sqrt{1+3sqrt{1+4sqrt{1+5sqrt{1+6sqrt{1+7sqrt{81}}}}}}$



                  $=sqrt{1+3sqrt{1+4sqrt{1+5sqrt{1+6sqrt{1+7sqrt{1+8sqrt{1+cdots}}}}}}}$



                  $5=sqrt{25}=sqrt{1+4sqrt{36}}=sqrt{1+4sqrt{1+5sqrt{49}}}=sqrt{1+4sqrt{1+5sqrt{1+6sqrt{64}}}}=sqrt{1+4sqrt{1+5sqrt{1+6sqrt{1+7sqrt{81}}}}}=sqrt{1+4sqrt{1+5sqrt{1+6sqrt{1+7sqrt{1+8sqrt{100}}}}}}$



                  $=sqrt{1+4sqrt{1+5sqrt{1+6sqrt{1+7sqrt{1+8sqrt{1+9sqrt{1+cdots}}}}}}}$



                  $vdots$



                  $n=sqrt{1+(n-1)sqrt{1+nsqrt{1+(n+1)sqrt{1+(n+2)sqrt{1+(n+3)sqrt{1+(n+4)sqrt{1+cdots}}}}}}}$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited 13 mins ago

























                  answered 39 mins ago









                  Okkes DulgerciOkkes Dulgerci

                  1193




                  1193






























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