Finding sum to infinity












3












$begingroup$


I am trying to find what this value will converge to
$$sum_{n = 1}^{ infty}frac{n^2}{n!}$$



I tried using the Taylor series for $e^x$ but couldn’t figure out how to manipulate it to get the above expression, can someone help me out.



Edit: I have seen the solution, the manipulation required didn’t come to me, is there any resource that you guys can tell me about where I can find/practice more questions like this?






calculus sequences-and-series taylor-expansion







share|cite|improve this question











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  • $begingroup$
    In general, $$ sum_{n=0}^{infty} frac{n^k}{n!} = B_k e, $$ where $B_k$ is the $k$-th Bell-number. This is called the Dobinski's formula. This can be computed by expanding $n^k$ into the sum of falling factorials as in Olivier Oloa's answer.
    $endgroup$
    – Sangchul Lee
    14 mins ago


















3












$begingroup$


I am trying to find what this value will converge to
$$sum_{n = 1}^{ infty}frac{n^2}{n!}$$



I tried using the Taylor series for $e^x$ but couldn’t figure out how to manipulate it to get the above expression, can someone help me out.



Edit: I have seen the solution, the manipulation required didn’t come to me, is there any resource that you guys can tell me about where I can find/practice more questions like this?






calculus sequences-and-series taylor-expansion







share|cite|improve this question











$endgroup$












  • $begingroup$
    In general, $$ sum_{n=0}^{infty} frac{n^k}{n!} = B_k e, $$ where $B_k$ is the $k$-th Bell-number. This is called the Dobinski's formula. This can be computed by expanding $n^k$ into the sum of falling factorials as in Olivier Oloa's answer.
    $endgroup$
    – Sangchul Lee
    14 mins ago
















3












3








3


1



$begingroup$


I am trying to find what this value will converge to
$$sum_{n = 1}^{ infty}frac{n^2}{n!}$$



I tried using the Taylor series for $e^x$ but couldn’t figure out how to manipulate it to get the above expression, can someone help me out.



Edit: I have seen the solution, the manipulation required didn’t come to me, is there any resource that you guys can tell me about where I can find/practice more questions like this?






calculus sequences-and-series taylor-expansion







share|cite|improve this question











$endgroup$




I am trying to find what this value will converge to
$$sum_{n = 1}^{ infty}frac{n^2}{n!}$$



I tried using the Taylor series for $e^x$ but couldn’t figure out how to manipulate it to get the above expression, can someone help me out.



Edit: I have seen the solution, the manipulation required didn’t come to me, is there any resource that you guys can tell me about where I can find/practice more questions like this?






calculus sequences-and-series taylor-expansion




calculus sequences-and-series taylor-expansion






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 4 mins ago







user601297

















asked 25 mins ago









user601297user601297

37019




37019












  • $begingroup$
    In general, $$ sum_{n=0}^{infty} frac{n^k}{n!} = B_k e, $$ where $B_k$ is the $k$-th Bell-number. This is called the Dobinski's formula. This can be computed by expanding $n^k$ into the sum of falling factorials as in Olivier Oloa's answer.
    $endgroup$
    – Sangchul Lee
    14 mins ago




















  • $begingroup$
    In general, $$ sum_{n=0}^{infty} frac{n^k}{n!} = B_k e, $$ where $B_k$ is the $k$-th Bell-number. This is called the Dobinski's formula. This can be computed by expanding $n^k$ into the sum of falling factorials as in Olivier Oloa's answer.
    $endgroup$
    – Sangchul Lee
    14 mins ago


















$begingroup$
In general, $$ sum_{n=0}^{infty} frac{n^k}{n!} = B_k e, $$ where $B_k$ is the $k$-th Bell-number. This is called the Dobinski's formula. This can be computed by expanding $n^k$ into the sum of falling factorials as in Olivier Oloa's answer.
$endgroup$
– Sangchul Lee
14 mins ago






$begingroup$
In general, $$ sum_{n=0}^{infty} frac{n^k}{n!} = B_k e, $$ where $B_k$ is the $k$-th Bell-number. This is called the Dobinski's formula. This can be computed by expanding $n^k$ into the sum of falling factorials as in Olivier Oloa's answer.
$endgroup$
– Sangchul Lee
14 mins ago












2 Answers
2






active

oldest

votes


















6












$begingroup$

One may write
begin{align}
sum_{n = 1}^{ infty}frac{n^2}{n!}&=sum_{n = 1}^{ infty}frac{n(n-1)+n}{n!}
\\&=sum_{n = 1}^{ infty}frac{n(n-1)}{n!}+sum_{n = 1}^{ infty}frac{n}{n!}
\\&=sum_{n = 2}^{ infty}frac{1}{(n-2)!}+sum_{n = 1}^{ infty}frac{1}{(n-1)!}
end{align}Can you take it from here?






share|cite|improve this answer









$endgroup$









  • 2




    $begingroup$
    Further hints: Since $e^1 = sum_{n=0}^{infty} frac{1}{n!}$, we have $$ sum_{n=2}^{infty} frac{1}{(n-2)!}= 1 + frac{1}{2} + frac{1}{3!} + .... = sum_{n=0}^{infty} frac{1}{n!} $$ and $$ sum_{n=1}^{infty} frac{1}{(n-1)!} = 1 + frac{1}{2} + frac{1}{3!} + .... = sum_{n=0}^{infty} frac{1}{n!} $$ So, ans: $boxed{ 2 mathrm{e} } $
    $endgroup$
    – Jimmy Sabater
    18 mins ago












  • $begingroup$
    Ok both expressions sum to $e$, i get it, thanks a lot
    $endgroup$
    – user601297
    11 mins ago



















2












$begingroup$

$$e^x-1=sum_{ngeq1}frac{x^n}{n!}$$
Taking $d/dx$ on both sides,
$$e^x=sum_{ngeq1}frac{n}{n!}x^{n-1}$$
Multiplying both sides by $x$,
$$xe^x=sum_{ngeq1}frac{n}{n!}x^n$$
Then taking $d/dx$ on both sides again,
$$(x+1)e^x=sum_{ngeq1}frac{n^2}{n!}x^{n-1}$$
Then plug in $x=1$:
$$sum_{ngeq1}frac{n^2}{n!}=2e$$






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Amazing, this is exactly what I was looking for
    $endgroup$
    – user601297
    6 mins ago










  • $begingroup$
    @user601297 you are very welcome :)
    $endgroup$
    – clathratus
    4 mins ago











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

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active

oldest

votes









6












$begingroup$

One may write
begin{align}
sum_{n = 1}^{ infty}frac{n^2}{n!}&=sum_{n = 1}^{ infty}frac{n(n-1)+n}{n!}
\\&=sum_{n = 1}^{ infty}frac{n(n-1)}{n!}+sum_{n = 1}^{ infty}frac{n}{n!}
\\&=sum_{n = 2}^{ infty}frac{1}{(n-2)!}+sum_{n = 1}^{ infty}frac{1}{(n-1)!}
end{align}Can you take it from here?






share|cite|improve this answer









$endgroup$









  • 2




    $begingroup$
    Further hints: Since $e^1 = sum_{n=0}^{infty} frac{1}{n!}$, we have $$ sum_{n=2}^{infty} frac{1}{(n-2)!}= 1 + frac{1}{2} + frac{1}{3!} + .... = sum_{n=0}^{infty} frac{1}{n!} $$ and $$ sum_{n=1}^{infty} frac{1}{(n-1)!} = 1 + frac{1}{2} + frac{1}{3!} + .... = sum_{n=0}^{infty} frac{1}{n!} $$ So, ans: $boxed{ 2 mathrm{e} } $
    $endgroup$
    – Jimmy Sabater
    18 mins ago












  • $begingroup$
    Ok both expressions sum to $e$, i get it, thanks a lot
    $endgroup$
    – user601297
    11 mins ago
















6












$begingroup$

One may write
begin{align}
sum_{n = 1}^{ infty}frac{n^2}{n!}&=sum_{n = 1}^{ infty}frac{n(n-1)+n}{n!}
\\&=sum_{n = 1}^{ infty}frac{n(n-1)}{n!}+sum_{n = 1}^{ infty}frac{n}{n!}
\\&=sum_{n = 2}^{ infty}frac{1}{(n-2)!}+sum_{n = 1}^{ infty}frac{1}{(n-1)!}
end{align}Can you take it from here?






share|cite|improve this answer









$endgroup$









  • 2




    $begingroup$
    Further hints: Since $e^1 = sum_{n=0}^{infty} frac{1}{n!}$, we have $$ sum_{n=2}^{infty} frac{1}{(n-2)!}= 1 + frac{1}{2} + frac{1}{3!} + .... = sum_{n=0}^{infty} frac{1}{n!} $$ and $$ sum_{n=1}^{infty} frac{1}{(n-1)!} = 1 + frac{1}{2} + frac{1}{3!} + .... = sum_{n=0}^{infty} frac{1}{n!} $$ So, ans: $boxed{ 2 mathrm{e} } $
    $endgroup$
    – Jimmy Sabater
    18 mins ago












  • $begingroup$
    Ok both expressions sum to $e$, i get it, thanks a lot
    $endgroup$
    – user601297
    11 mins ago














6












6








6





$begingroup$

One may write
begin{align}
sum_{n = 1}^{ infty}frac{n^2}{n!}&=sum_{n = 1}^{ infty}frac{n(n-1)+n}{n!}
\\&=sum_{n = 1}^{ infty}frac{n(n-1)}{n!}+sum_{n = 1}^{ infty}frac{n}{n!}
\\&=sum_{n = 2}^{ infty}frac{1}{(n-2)!}+sum_{n = 1}^{ infty}frac{1}{(n-1)!}
end{align}Can you take it from here?






share|cite|improve this answer









$endgroup$



One may write
begin{align}
sum_{n = 1}^{ infty}frac{n^2}{n!}&=sum_{n = 1}^{ infty}frac{n(n-1)+n}{n!}
\\&=sum_{n = 1}^{ infty}frac{n(n-1)}{n!}+sum_{n = 1}^{ infty}frac{n}{n!}
\\&=sum_{n = 2}^{ infty}frac{1}{(n-2)!}+sum_{n = 1}^{ infty}frac{1}{(n-1)!}
end{align}Can you take it from here?







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 20 mins ago









Olivier OloaOlivier Oloa

108k17176293




108k17176293








  • 2




    $begingroup$
    Further hints: Since $e^1 = sum_{n=0}^{infty} frac{1}{n!}$, we have $$ sum_{n=2}^{infty} frac{1}{(n-2)!}= 1 + frac{1}{2} + frac{1}{3!} + .... = sum_{n=0}^{infty} frac{1}{n!} $$ and $$ sum_{n=1}^{infty} frac{1}{(n-1)!} = 1 + frac{1}{2} + frac{1}{3!} + .... = sum_{n=0}^{infty} frac{1}{n!} $$ So, ans: $boxed{ 2 mathrm{e} } $
    $endgroup$
    – Jimmy Sabater
    18 mins ago












  • $begingroup$
    Ok both expressions sum to $e$, i get it, thanks a lot
    $endgroup$
    – user601297
    11 mins ago














  • 2




    $begingroup$
    Further hints: Since $e^1 = sum_{n=0}^{infty} frac{1}{n!}$, we have $$ sum_{n=2}^{infty} frac{1}{(n-2)!}= 1 + frac{1}{2} + frac{1}{3!} + .... = sum_{n=0}^{infty} frac{1}{n!} $$ and $$ sum_{n=1}^{infty} frac{1}{(n-1)!} = 1 + frac{1}{2} + frac{1}{3!} + .... = sum_{n=0}^{infty} frac{1}{n!} $$ So, ans: $boxed{ 2 mathrm{e} } $
    $endgroup$
    – Jimmy Sabater
    18 mins ago












  • $begingroup$
    Ok both expressions sum to $e$, i get it, thanks a lot
    $endgroup$
    – user601297
    11 mins ago








2




2




$begingroup$
Further hints: Since $e^1 = sum_{n=0}^{infty} frac{1}{n!}$, we have $$ sum_{n=2}^{infty} frac{1}{(n-2)!}= 1 + frac{1}{2} + frac{1}{3!} + .... = sum_{n=0}^{infty} frac{1}{n!} $$ and $$ sum_{n=1}^{infty} frac{1}{(n-1)!} = 1 + frac{1}{2} + frac{1}{3!} + .... = sum_{n=0}^{infty} frac{1}{n!} $$ So, ans: $boxed{ 2 mathrm{e} } $
$endgroup$
– Jimmy Sabater
18 mins ago






$begingroup$
Further hints: Since $e^1 = sum_{n=0}^{infty} frac{1}{n!}$, we have $$ sum_{n=2}^{infty} frac{1}{(n-2)!}= 1 + frac{1}{2} + frac{1}{3!} + .... = sum_{n=0}^{infty} frac{1}{n!} $$ and $$ sum_{n=1}^{infty} frac{1}{(n-1)!} = 1 + frac{1}{2} + frac{1}{3!} + .... = sum_{n=0}^{infty} frac{1}{n!} $$ So, ans: $boxed{ 2 mathrm{e} } $
$endgroup$
– Jimmy Sabater
18 mins ago














$begingroup$
Ok both expressions sum to $e$, i get it, thanks a lot
$endgroup$
– user601297
11 mins ago




$begingroup$
Ok both expressions sum to $e$, i get it, thanks a lot
$endgroup$
– user601297
11 mins ago











2












$begingroup$

$$e^x-1=sum_{ngeq1}frac{x^n}{n!}$$
Taking $d/dx$ on both sides,
$$e^x=sum_{ngeq1}frac{n}{n!}x^{n-1}$$
Multiplying both sides by $x$,
$$xe^x=sum_{ngeq1}frac{n}{n!}x^n$$
Then taking $d/dx$ on both sides again,
$$(x+1)e^x=sum_{ngeq1}frac{n^2}{n!}x^{n-1}$$
Then plug in $x=1$:
$$sum_{ngeq1}frac{n^2}{n!}=2e$$






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Amazing, this is exactly what I was looking for
    $endgroup$
    – user601297
    6 mins ago










  • $begingroup$
    @user601297 you are very welcome :)
    $endgroup$
    – clathratus
    4 mins ago
















2












$begingroup$

$$e^x-1=sum_{ngeq1}frac{x^n}{n!}$$
Taking $d/dx$ on both sides,
$$e^x=sum_{ngeq1}frac{n}{n!}x^{n-1}$$
Multiplying both sides by $x$,
$$xe^x=sum_{ngeq1}frac{n}{n!}x^n$$
Then taking $d/dx$ on both sides again,
$$(x+1)e^x=sum_{ngeq1}frac{n^2}{n!}x^{n-1}$$
Then plug in $x=1$:
$$sum_{ngeq1}frac{n^2}{n!}=2e$$






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Amazing, this is exactly what I was looking for
    $endgroup$
    – user601297
    6 mins ago










  • $begingroup$
    @user601297 you are very welcome :)
    $endgroup$
    – clathratus
    4 mins ago














2












2








2





$begingroup$

$$e^x-1=sum_{ngeq1}frac{x^n}{n!}$$
Taking $d/dx$ on both sides,
$$e^x=sum_{ngeq1}frac{n}{n!}x^{n-1}$$
Multiplying both sides by $x$,
$$xe^x=sum_{ngeq1}frac{n}{n!}x^n$$
Then taking $d/dx$ on both sides again,
$$(x+1)e^x=sum_{ngeq1}frac{n^2}{n!}x^{n-1}$$
Then plug in $x=1$:
$$sum_{ngeq1}frac{n^2}{n!}=2e$$






share|cite|improve this answer









$endgroup$



$$e^x-1=sum_{ngeq1}frac{x^n}{n!}$$
Taking $d/dx$ on both sides,
$$e^x=sum_{ngeq1}frac{n}{n!}x^{n-1}$$
Multiplying both sides by $x$,
$$xe^x=sum_{ngeq1}frac{n}{n!}x^n$$
Then taking $d/dx$ on both sides again,
$$(x+1)e^x=sum_{ngeq1}frac{n^2}{n!}x^{n-1}$$
Then plug in $x=1$:
$$sum_{ngeq1}frac{n^2}{n!}=2e$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 10 mins ago









clathratusclathratus

3,651332




3,651332








  • 1




    $begingroup$
    Amazing, this is exactly what I was looking for
    $endgroup$
    – user601297
    6 mins ago










  • $begingroup$
    @user601297 you are very welcome :)
    $endgroup$
    – clathratus
    4 mins ago














  • 1




    $begingroup$
    Amazing, this is exactly what I was looking for
    $endgroup$
    – user601297
    6 mins ago










  • $begingroup$
    @user601297 you are very welcome :)
    $endgroup$
    – clathratus
    4 mins ago








1




1




$begingroup$
Amazing, this is exactly what I was looking for
$endgroup$
– user601297
6 mins ago




$begingroup$
Amazing, this is exactly what I was looking for
$endgroup$
– user601297
6 mins ago












$begingroup$
@user601297 you are very welcome :)
$endgroup$
– clathratus
4 mins ago




$begingroup$
@user601297 you are very welcome :)
$endgroup$
– clathratus
4 mins ago


















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