Making sure if it is Cauchy
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In my real analysis exam I had a problem in which I proved that $|x_{n+1} - x_n|lt frac 1{a^n}$ for all natural numbers $n$ and for all positive number $alt 1$ then $(x_n)$ is a Cauchy sequence.
This was solved successfully but the question is if $|x_{n+1} - x_n|lt frac 1n$ does that mean $(x_n)$ is Cauchy? Well my answer was yes because I could write this in the form of the first one, but now I am somehow confused with what I have answered since $1/n$ is a sequence of $n$ so maybe the answer is not necessarily true... Can you please provide me with the correct answer for this question?
real-analysis cauchy-sequences
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up vote
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favorite
In my real analysis exam I had a problem in which I proved that $|x_{n+1} - x_n|lt frac 1{a^n}$ for all natural numbers $n$ and for all positive number $alt 1$ then $(x_n)$ is a Cauchy sequence.
This was solved successfully but the question is if $|x_{n+1} - x_n|lt frac 1n$ does that mean $(x_n)$ is Cauchy? Well my answer was yes because I could write this in the form of the first one, but now I am somehow confused with what I have answered since $1/n$ is a sequence of $n$ so maybe the answer is not necessarily true... Can you please provide me with the correct answer for this question?
real-analysis cauchy-sequences
slow moving thread -_-
– Matt A Pelto
1 hour ago
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
In my real analysis exam I had a problem in which I proved that $|x_{n+1} - x_n|lt frac 1{a^n}$ for all natural numbers $n$ and for all positive number $alt 1$ then $(x_n)$ is a Cauchy sequence.
This was solved successfully but the question is if $|x_{n+1} - x_n|lt frac 1n$ does that mean $(x_n)$ is Cauchy? Well my answer was yes because I could write this in the form of the first one, but now I am somehow confused with what I have answered since $1/n$ is a sequence of $n$ so maybe the answer is not necessarily true... Can you please provide me with the correct answer for this question?
real-analysis cauchy-sequences
In my real analysis exam I had a problem in which I proved that $|x_{n+1} - x_n|lt frac 1{a^n}$ for all natural numbers $n$ and for all positive number $alt 1$ then $(x_n)$ is a Cauchy sequence.
This was solved successfully but the question is if $|x_{n+1} - x_n|lt frac 1n$ does that mean $(x_n)$ is Cauchy? Well my answer was yes because I could write this in the form of the first one, but now I am somehow confused with what I have answered since $1/n$ is a sequence of $n$ so maybe the answer is not necessarily true... Can you please provide me with the correct answer for this question?
real-analysis cauchy-sequences
real-analysis cauchy-sequences
edited 1 hour ago
amWhy
191k27223437
191k27223437
asked 1 hour ago
user7857462
311
311
slow moving thread -_-
– Matt A Pelto
1 hour ago
add a comment |
slow moving thread -_-
– Matt A Pelto
1 hour ago
slow moving thread -_-
– Matt A Pelto
1 hour ago
slow moving thread -_-
– Matt A Pelto
1 hour ago
add a comment |
6 Answers
6
active
oldest
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up vote
3
down vote
No, $lvert x_{n+1}-x_nrvert < 1/n$ does not imply that $(x_n)_{nin mathbb{N}}$ is Cauchy. Consider $x_n = sum_{k=1}^n 1/2k$, which does not converge.
$1/2n$ converges, so it is Cauchy!
– Bernard
1 hour ago
Yes, I was missing the sum.
– Michael Lee
1 hour ago
add a comment |
up vote
3
down vote
Consider the harmonic series: $sum_{n=1}^{infty}frac 1n$.
$mid a_{n+1}-a_nmid=frac1{n+1}ltfrac1n$.
But it diverges.
This is the difference between $1/n$ and $1/(n+1)$, not between two terms of the sequence of partial sums. The sequence $(1/n)_{nin mathbb{N}}$ is indeed Cauchy.
– Michael Lee
1 hour ago
Yes. I'm referring to the sequence of partial sums of the series. It's not Cauchy because it doesn't converge. I was addressing the OP's example. This shows that $mid a_{n+1}-a_nmid$ can be less than $frac1n$, but the sequence can still not be Cauchy.
– Chris Custer
54 mins ago
The difference between partial sums is $1/(n+1)$, not $1/n(n+1)$. You add $1/(n+1)$ to get from the $n$th partial sum to the $(n+1)$th.
– Michael Lee
54 mins ago
Oh yeah. My mistake.
– Chris Custer
51 mins ago
add a comment |
up vote
2
down vote
Take $x_n=1+frac 1 2+cdots+frac 1 n$. This is not Cauchy because the harmonic series $1+frac 1 2+cdots$ is divergent.
add a comment |
up vote
1
down vote
This would imply that any series with a general term which tends to $0$ is convergent. This is false (except for $p$-adic numbers…).
add a comment |
up vote
1
down vote
For $n=1,2,ldots$ define $x_n:=1^{-1}+cdots+n^{-1}$. Notice $|x_{n+1}-x_n|=frac{1}{n+1}$ but the sequence ${x_n}$ is not Cauchy as its terms are just the partial sums of a series that is known to not converge and $mathbb{R}$ is complete.
add a comment |
up vote
1
down vote
This kind of thing works only if you can show $|x_{n + 1} - x_{n}| < a_n$ where $sum_{k = 0}^infty a_k < infty$ because, if this condition holds,
begin{align}
|x_{n} - x_{n + m}| &= |x_{n} - x_{n + 1} + x_{n + 1} - x_{n + 2} + x_{n + 2} - cdots + x_{n + m - 1} - x_{n + m}| \
&le |x_{n} - x_{n + 1}| + cdots + |x_{n + m - 1} - x_{n + m}| \
&le a_n + a_{n + 1} + dots + a_{n + m - 1} \
&le sum_{k = n}^infty a_k
end{align}
Now convergence of $sum a_k$ to $A$ means that for any $varepsilon > 0$ there exists $N$ such that for all $n ge N$,
$$ left| A - sum_{k = 0}^{n - 1} a_k right| = sum_{k = n}^infty a_k < varepsilon $$
Comparing this with the above, we have for every $n ge N$ and every $m ge 0$,
$$ |x_n - x_{n + m}| < varepsilon $$
Which means the sequence $(x_n)$ is Cauchy.
If the bound on $|x_{n + 1} - x_n|$ does not converge as a series, you need to use a different trick.
1
+1 for the most informative response and using varepsilon...no effort has been spared here. I will however nitpick at the last sentence by noting that a different trick may or may not still work, depending on whether the sequence is indeed Cauchy which I presume you know but just add as clarification for people such as the question asker. An example where a different trick might apply is the sequence of partial sums of the alternating series $sum_{n=1}^infty frac{(-1)^{n+1}}{n}$.
– Matt A Pelto
34 mins ago
add a comment |
6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
No, $lvert x_{n+1}-x_nrvert < 1/n$ does not imply that $(x_n)_{nin mathbb{N}}$ is Cauchy. Consider $x_n = sum_{k=1}^n 1/2k$, which does not converge.
$1/2n$ converges, so it is Cauchy!
– Bernard
1 hour ago
Yes, I was missing the sum.
– Michael Lee
1 hour ago
add a comment |
up vote
3
down vote
No, $lvert x_{n+1}-x_nrvert < 1/n$ does not imply that $(x_n)_{nin mathbb{N}}$ is Cauchy. Consider $x_n = sum_{k=1}^n 1/2k$, which does not converge.
$1/2n$ converges, so it is Cauchy!
– Bernard
1 hour ago
Yes, I was missing the sum.
– Michael Lee
1 hour ago
add a comment |
up vote
3
down vote
up vote
3
down vote
No, $lvert x_{n+1}-x_nrvert < 1/n$ does not imply that $(x_n)_{nin mathbb{N}}$ is Cauchy. Consider $x_n = sum_{k=1}^n 1/2k$, which does not converge.
No, $lvert x_{n+1}-x_nrvert < 1/n$ does not imply that $(x_n)_{nin mathbb{N}}$ is Cauchy. Consider $x_n = sum_{k=1}^n 1/2k$, which does not converge.
answered 1 hour ago
Michael Lee
4,6901929
4,6901929
$1/2n$ converges, so it is Cauchy!
– Bernard
1 hour ago
Yes, I was missing the sum.
– Michael Lee
1 hour ago
add a comment |
$1/2n$ converges, so it is Cauchy!
– Bernard
1 hour ago
Yes, I was missing the sum.
– Michael Lee
1 hour ago
$1/2n$ converges, so it is Cauchy!
– Bernard
1 hour ago
$1/2n$ converges, so it is Cauchy!
– Bernard
1 hour ago
Yes, I was missing the sum.
– Michael Lee
1 hour ago
Yes, I was missing the sum.
– Michael Lee
1 hour ago
add a comment |
up vote
3
down vote
Consider the harmonic series: $sum_{n=1}^{infty}frac 1n$.
$mid a_{n+1}-a_nmid=frac1{n+1}ltfrac1n$.
But it diverges.
This is the difference between $1/n$ and $1/(n+1)$, not between two terms of the sequence of partial sums. The sequence $(1/n)_{nin mathbb{N}}$ is indeed Cauchy.
– Michael Lee
1 hour ago
Yes. I'm referring to the sequence of partial sums of the series. It's not Cauchy because it doesn't converge. I was addressing the OP's example. This shows that $mid a_{n+1}-a_nmid$ can be less than $frac1n$, but the sequence can still not be Cauchy.
– Chris Custer
54 mins ago
The difference between partial sums is $1/(n+1)$, not $1/n(n+1)$. You add $1/(n+1)$ to get from the $n$th partial sum to the $(n+1)$th.
– Michael Lee
54 mins ago
Oh yeah. My mistake.
– Chris Custer
51 mins ago
add a comment |
up vote
3
down vote
Consider the harmonic series: $sum_{n=1}^{infty}frac 1n$.
$mid a_{n+1}-a_nmid=frac1{n+1}ltfrac1n$.
But it diverges.
This is the difference between $1/n$ and $1/(n+1)$, not between two terms of the sequence of partial sums. The sequence $(1/n)_{nin mathbb{N}}$ is indeed Cauchy.
– Michael Lee
1 hour ago
Yes. I'm referring to the sequence of partial sums of the series. It's not Cauchy because it doesn't converge. I was addressing the OP's example. This shows that $mid a_{n+1}-a_nmid$ can be less than $frac1n$, but the sequence can still not be Cauchy.
– Chris Custer
54 mins ago
The difference between partial sums is $1/(n+1)$, not $1/n(n+1)$. You add $1/(n+1)$ to get from the $n$th partial sum to the $(n+1)$th.
– Michael Lee
54 mins ago
Oh yeah. My mistake.
– Chris Custer
51 mins ago
add a comment |
up vote
3
down vote
up vote
3
down vote
Consider the harmonic series: $sum_{n=1}^{infty}frac 1n$.
$mid a_{n+1}-a_nmid=frac1{n+1}ltfrac1n$.
But it diverges.
Consider the harmonic series: $sum_{n=1}^{infty}frac 1n$.
$mid a_{n+1}-a_nmid=frac1{n+1}ltfrac1n$.
But it diverges.
edited 45 mins ago
answered 1 hour ago
Chris Custer
8,9643623
8,9643623
This is the difference between $1/n$ and $1/(n+1)$, not between two terms of the sequence of partial sums. The sequence $(1/n)_{nin mathbb{N}}$ is indeed Cauchy.
– Michael Lee
1 hour ago
Yes. I'm referring to the sequence of partial sums of the series. It's not Cauchy because it doesn't converge. I was addressing the OP's example. This shows that $mid a_{n+1}-a_nmid$ can be less than $frac1n$, but the sequence can still not be Cauchy.
– Chris Custer
54 mins ago
The difference between partial sums is $1/(n+1)$, not $1/n(n+1)$. You add $1/(n+1)$ to get from the $n$th partial sum to the $(n+1)$th.
– Michael Lee
54 mins ago
Oh yeah. My mistake.
– Chris Custer
51 mins ago
add a comment |
This is the difference between $1/n$ and $1/(n+1)$, not between two terms of the sequence of partial sums. The sequence $(1/n)_{nin mathbb{N}}$ is indeed Cauchy.
– Michael Lee
1 hour ago
Yes. I'm referring to the sequence of partial sums of the series. It's not Cauchy because it doesn't converge. I was addressing the OP's example. This shows that $mid a_{n+1}-a_nmid$ can be less than $frac1n$, but the sequence can still not be Cauchy.
– Chris Custer
54 mins ago
The difference between partial sums is $1/(n+1)$, not $1/n(n+1)$. You add $1/(n+1)$ to get from the $n$th partial sum to the $(n+1)$th.
– Michael Lee
54 mins ago
Oh yeah. My mistake.
– Chris Custer
51 mins ago
This is the difference between $1/n$ and $1/(n+1)$, not between two terms of the sequence of partial sums. The sequence $(1/n)_{nin mathbb{N}}$ is indeed Cauchy.
– Michael Lee
1 hour ago
This is the difference between $1/n$ and $1/(n+1)$, not between two terms of the sequence of partial sums. The sequence $(1/n)_{nin mathbb{N}}$ is indeed Cauchy.
– Michael Lee
1 hour ago
Yes. I'm referring to the sequence of partial sums of the series. It's not Cauchy because it doesn't converge. I was addressing the OP's example. This shows that $mid a_{n+1}-a_nmid$ can be less than $frac1n$, but the sequence can still not be Cauchy.
– Chris Custer
54 mins ago
Yes. I'm referring to the sequence of partial sums of the series. It's not Cauchy because it doesn't converge. I was addressing the OP's example. This shows that $mid a_{n+1}-a_nmid$ can be less than $frac1n$, but the sequence can still not be Cauchy.
– Chris Custer
54 mins ago
The difference between partial sums is $1/(n+1)$, not $1/n(n+1)$. You add $1/(n+1)$ to get from the $n$th partial sum to the $(n+1)$th.
– Michael Lee
54 mins ago
The difference between partial sums is $1/(n+1)$, not $1/n(n+1)$. You add $1/(n+1)$ to get from the $n$th partial sum to the $(n+1)$th.
– Michael Lee
54 mins ago
Oh yeah. My mistake.
– Chris Custer
51 mins ago
Oh yeah. My mistake.
– Chris Custer
51 mins ago
add a comment |
up vote
2
down vote
Take $x_n=1+frac 1 2+cdots+frac 1 n$. This is not Cauchy because the harmonic series $1+frac 1 2+cdots$ is divergent.
add a comment |
up vote
2
down vote
Take $x_n=1+frac 1 2+cdots+frac 1 n$. This is not Cauchy because the harmonic series $1+frac 1 2+cdots$ is divergent.
add a comment |
up vote
2
down vote
up vote
2
down vote
Take $x_n=1+frac 1 2+cdots+frac 1 n$. This is not Cauchy because the harmonic series $1+frac 1 2+cdots$ is divergent.
Take $x_n=1+frac 1 2+cdots+frac 1 n$. This is not Cauchy because the harmonic series $1+frac 1 2+cdots$ is divergent.
answered 1 hour ago
Kavi Rama Murthy
41.6k31751
41.6k31751
add a comment |
add a comment |
up vote
1
down vote
This would imply that any series with a general term which tends to $0$ is convergent. This is false (except for $p$-adic numbers…).
add a comment |
up vote
1
down vote
This would imply that any series with a general term which tends to $0$ is convergent. This is false (except for $p$-adic numbers…).
add a comment |
up vote
1
down vote
up vote
1
down vote
This would imply that any series with a general term which tends to $0$ is convergent. This is false (except for $p$-adic numbers…).
This would imply that any series with a general term which tends to $0$ is convergent. This is false (except for $p$-adic numbers…).
answered 1 hour ago
Bernard
116k637107
116k637107
add a comment |
add a comment |
up vote
1
down vote
For $n=1,2,ldots$ define $x_n:=1^{-1}+cdots+n^{-1}$. Notice $|x_{n+1}-x_n|=frac{1}{n+1}$ but the sequence ${x_n}$ is not Cauchy as its terms are just the partial sums of a series that is known to not converge and $mathbb{R}$ is complete.
add a comment |
up vote
1
down vote
For $n=1,2,ldots$ define $x_n:=1^{-1}+cdots+n^{-1}$. Notice $|x_{n+1}-x_n|=frac{1}{n+1}$ but the sequence ${x_n}$ is not Cauchy as its terms are just the partial sums of a series that is known to not converge and $mathbb{R}$ is complete.
add a comment |
up vote
1
down vote
up vote
1
down vote
For $n=1,2,ldots$ define $x_n:=1^{-1}+cdots+n^{-1}$. Notice $|x_{n+1}-x_n|=frac{1}{n+1}$ but the sequence ${x_n}$ is not Cauchy as its terms are just the partial sums of a series that is known to not converge and $mathbb{R}$ is complete.
For $n=1,2,ldots$ define $x_n:=1^{-1}+cdots+n^{-1}$. Notice $|x_{n+1}-x_n|=frac{1}{n+1}$ but the sequence ${x_n}$ is not Cauchy as its terms are just the partial sums of a series that is known to not converge and $mathbb{R}$ is complete.
answered 1 hour ago
Matt A Pelto
2,095519
2,095519
add a comment |
add a comment |
up vote
1
down vote
This kind of thing works only if you can show $|x_{n + 1} - x_{n}| < a_n$ where $sum_{k = 0}^infty a_k < infty$ because, if this condition holds,
begin{align}
|x_{n} - x_{n + m}| &= |x_{n} - x_{n + 1} + x_{n + 1} - x_{n + 2} + x_{n + 2} - cdots + x_{n + m - 1} - x_{n + m}| \
&le |x_{n} - x_{n + 1}| + cdots + |x_{n + m - 1} - x_{n + m}| \
&le a_n + a_{n + 1} + dots + a_{n + m - 1} \
&le sum_{k = n}^infty a_k
end{align}
Now convergence of $sum a_k$ to $A$ means that for any $varepsilon > 0$ there exists $N$ such that for all $n ge N$,
$$ left| A - sum_{k = 0}^{n - 1} a_k right| = sum_{k = n}^infty a_k < varepsilon $$
Comparing this with the above, we have for every $n ge N$ and every $m ge 0$,
$$ |x_n - x_{n + m}| < varepsilon $$
Which means the sequence $(x_n)$ is Cauchy.
If the bound on $|x_{n + 1} - x_n|$ does not converge as a series, you need to use a different trick.
1
+1 for the most informative response and using varepsilon...no effort has been spared here. I will however nitpick at the last sentence by noting that a different trick may or may not still work, depending on whether the sequence is indeed Cauchy which I presume you know but just add as clarification for people such as the question asker. An example where a different trick might apply is the sequence of partial sums of the alternating series $sum_{n=1}^infty frac{(-1)^{n+1}}{n}$.
– Matt A Pelto
34 mins ago
add a comment |
up vote
1
down vote
This kind of thing works only if you can show $|x_{n + 1} - x_{n}| < a_n$ where $sum_{k = 0}^infty a_k < infty$ because, if this condition holds,
begin{align}
|x_{n} - x_{n + m}| &= |x_{n} - x_{n + 1} + x_{n + 1} - x_{n + 2} + x_{n + 2} - cdots + x_{n + m - 1} - x_{n + m}| \
&le |x_{n} - x_{n + 1}| + cdots + |x_{n + m - 1} - x_{n + m}| \
&le a_n + a_{n + 1} + dots + a_{n + m - 1} \
&le sum_{k = n}^infty a_k
end{align}
Now convergence of $sum a_k$ to $A$ means that for any $varepsilon > 0$ there exists $N$ such that for all $n ge N$,
$$ left| A - sum_{k = 0}^{n - 1} a_k right| = sum_{k = n}^infty a_k < varepsilon $$
Comparing this with the above, we have for every $n ge N$ and every $m ge 0$,
$$ |x_n - x_{n + m}| < varepsilon $$
Which means the sequence $(x_n)$ is Cauchy.
If the bound on $|x_{n + 1} - x_n|$ does not converge as a series, you need to use a different trick.
1
+1 for the most informative response and using varepsilon...no effort has been spared here. I will however nitpick at the last sentence by noting that a different trick may or may not still work, depending on whether the sequence is indeed Cauchy which I presume you know but just add as clarification for people such as the question asker. An example where a different trick might apply is the sequence of partial sums of the alternating series $sum_{n=1}^infty frac{(-1)^{n+1}}{n}$.
– Matt A Pelto
34 mins ago
add a comment |
up vote
1
down vote
up vote
1
down vote
This kind of thing works only if you can show $|x_{n + 1} - x_{n}| < a_n$ where $sum_{k = 0}^infty a_k < infty$ because, if this condition holds,
begin{align}
|x_{n} - x_{n + m}| &= |x_{n} - x_{n + 1} + x_{n + 1} - x_{n + 2} + x_{n + 2} - cdots + x_{n + m - 1} - x_{n + m}| \
&le |x_{n} - x_{n + 1}| + cdots + |x_{n + m - 1} - x_{n + m}| \
&le a_n + a_{n + 1} + dots + a_{n + m - 1} \
&le sum_{k = n}^infty a_k
end{align}
Now convergence of $sum a_k$ to $A$ means that for any $varepsilon > 0$ there exists $N$ such that for all $n ge N$,
$$ left| A - sum_{k = 0}^{n - 1} a_k right| = sum_{k = n}^infty a_k < varepsilon $$
Comparing this with the above, we have for every $n ge N$ and every $m ge 0$,
$$ |x_n - x_{n + m}| < varepsilon $$
Which means the sequence $(x_n)$ is Cauchy.
If the bound on $|x_{n + 1} - x_n|$ does not converge as a series, you need to use a different trick.
This kind of thing works only if you can show $|x_{n + 1} - x_{n}| < a_n$ where $sum_{k = 0}^infty a_k < infty$ because, if this condition holds,
begin{align}
|x_{n} - x_{n + m}| &= |x_{n} - x_{n + 1} + x_{n + 1} - x_{n + 2} + x_{n + 2} - cdots + x_{n + m - 1} - x_{n + m}| \
&le |x_{n} - x_{n + 1}| + cdots + |x_{n + m - 1} - x_{n + m}| \
&le a_n + a_{n + 1} + dots + a_{n + m - 1} \
&le sum_{k = n}^infty a_k
end{align}
Now convergence of $sum a_k$ to $A$ means that for any $varepsilon > 0$ there exists $N$ such that for all $n ge N$,
$$ left| A - sum_{k = 0}^{n - 1} a_k right| = sum_{k = n}^infty a_k < varepsilon $$
Comparing this with the above, we have for every $n ge N$ and every $m ge 0$,
$$ |x_n - x_{n + m}| < varepsilon $$
Which means the sequence $(x_n)$ is Cauchy.
If the bound on $|x_{n + 1} - x_n|$ does not converge as a series, you need to use a different trick.
answered 1 hour ago
Trevor Gunn
13.8k32045
13.8k32045
1
+1 for the most informative response and using varepsilon...no effort has been spared here. I will however nitpick at the last sentence by noting that a different trick may or may not still work, depending on whether the sequence is indeed Cauchy which I presume you know but just add as clarification for people such as the question asker. An example where a different trick might apply is the sequence of partial sums of the alternating series $sum_{n=1}^infty frac{(-1)^{n+1}}{n}$.
– Matt A Pelto
34 mins ago
add a comment |
1
+1 for the most informative response and using varepsilon...no effort has been spared here. I will however nitpick at the last sentence by noting that a different trick may or may not still work, depending on whether the sequence is indeed Cauchy which I presume you know but just add as clarification for people such as the question asker. An example where a different trick might apply is the sequence of partial sums of the alternating series $sum_{n=1}^infty frac{(-1)^{n+1}}{n}$.
– Matt A Pelto
34 mins ago
1
1
+1 for the most informative response and using varepsilon...no effort has been spared here. I will however nitpick at the last sentence by noting that a different trick may or may not still work, depending on whether the sequence is indeed Cauchy which I presume you know but just add as clarification for people such as the question asker. An example where a different trick might apply is the sequence of partial sums of the alternating series $sum_{n=1}^infty frac{(-1)^{n+1}}{n}$.
– Matt A Pelto
34 mins ago
+1 for the most informative response and using varepsilon...no effort has been spared here. I will however nitpick at the last sentence by noting that a different trick may or may not still work, depending on whether the sequence is indeed Cauchy which I presume you know but just add as clarification for people such as the question asker. An example where a different trick might apply is the sequence of partial sums of the alternating series $sum_{n=1}^infty frac{(-1)^{n+1}}{n}$.
– Matt A Pelto
34 mins ago
add a comment |
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slow moving thread -_-
– Matt A Pelto
1 hour ago