Finding sum to infinity [duplicate]












5












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This question already has an answer here:




  • What's the value of $sumlimits_{k=1}^{infty}frac{k^2}{k!}$?

    5 answers




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?










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marked as duplicate by lab bhattacharjee calculus
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40 mins ago


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.















  • $begingroup$
    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
    2 hours ago












  • $begingroup$
    If $R>0$ and a power series $sum_{j=0}^{infty}a_jx^j$ converges to $f(x)$ for each $xin (-R,R)$ then $f$ is differentiable on $(-R,R)$ and $sum_{j=1}^{infty}ja_jx^{j-1} =f'(x)$ for all $(-R,R).$ Notice also that the previous sentence now applies with $f$ replaced by $f'$. For example, for $|x|<2$ let $f(x) =1/(1-x)=sum_{j=0}^{infty}x^j .$ Then $sum_{j=1}^{infty}jx^{j-1}=f'(x)=1/(1-x)^2,$ and $sum_{j=2}^{infty}j(j-1)x^{j-2}=f''(x)=2/(1-x)^3.$
    $endgroup$
    – DanielWainfleet
    1 hour ago












  • $begingroup$
    See : math.stackexchange.com/questions/1711318/…
    $endgroup$
    – lab bhattacharjee
    40 mins ago
















5












$begingroup$



This question already has an answer here:




  • What's the value of $sumlimits_{k=1}^{infty}frac{k^2}{k!}$?

    5 answers




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?










share|cite|improve this question











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40 mins ago


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.















  • $begingroup$
    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
    2 hours ago












  • $begingroup$
    If $R>0$ and a power series $sum_{j=0}^{infty}a_jx^j$ converges to $f(x)$ for each $xin (-R,R)$ then $f$ is differentiable on $(-R,R)$ and $sum_{j=1}^{infty}ja_jx^{j-1} =f'(x)$ for all $(-R,R).$ Notice also that the previous sentence now applies with $f$ replaced by $f'$. For example, for $|x|<2$ let $f(x) =1/(1-x)=sum_{j=0}^{infty}x^j .$ Then $sum_{j=1}^{infty}jx^{j-1}=f'(x)=1/(1-x)^2,$ and $sum_{j=2}^{infty}j(j-1)x^{j-2}=f''(x)=2/(1-x)^3.$
    $endgroup$
    – DanielWainfleet
    1 hour ago












  • $begingroup$
    See : math.stackexchange.com/questions/1711318/…
    $endgroup$
    – lab bhattacharjee
    40 mins ago














5












5








5


1



$begingroup$



This question already has an answer here:




  • What's the value of $sumlimits_{k=1}^{infty}frac{k^2}{k!}$?

    5 answers




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?










share|cite|improve this question











$endgroup$





This question already has an answer here:




  • What's the value of $sumlimits_{k=1}^{infty}frac{k^2}{k!}$?

    5 answers




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?





This question already has an answer here:




  • What's the value of $sumlimits_{k=1}^{infty}frac{k^2}{k!}$?

    5 answers








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 1 hour ago







user601297

















asked 2 hours ago









user601297user601297

38319




38319




marked as duplicate by lab bhattacharjee calculus
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40 mins ago


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.






marked as duplicate by lab bhattacharjee calculus
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40 mins ago


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • $begingroup$
    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
    2 hours ago












  • $begingroup$
    If $R>0$ and a power series $sum_{j=0}^{infty}a_jx^j$ converges to $f(x)$ for each $xin (-R,R)$ then $f$ is differentiable on $(-R,R)$ and $sum_{j=1}^{infty}ja_jx^{j-1} =f'(x)$ for all $(-R,R).$ Notice also that the previous sentence now applies with $f$ replaced by $f'$. For example, for $|x|<2$ let $f(x) =1/(1-x)=sum_{j=0}^{infty}x^j .$ Then $sum_{j=1}^{infty}jx^{j-1}=f'(x)=1/(1-x)^2,$ and $sum_{j=2}^{infty}j(j-1)x^{j-2}=f''(x)=2/(1-x)^3.$
    $endgroup$
    – DanielWainfleet
    1 hour ago












  • $begingroup$
    See : math.stackexchange.com/questions/1711318/…
    $endgroup$
    – lab bhattacharjee
    40 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
    2 hours ago












  • $begingroup$
    If $R>0$ and a power series $sum_{j=0}^{infty}a_jx^j$ converges to $f(x)$ for each $xin (-R,R)$ then $f$ is differentiable on $(-R,R)$ and $sum_{j=1}^{infty}ja_jx^{j-1} =f'(x)$ for all $(-R,R).$ Notice also that the previous sentence now applies with $f$ replaced by $f'$. For example, for $|x|<2$ let $f(x) =1/(1-x)=sum_{j=0}^{infty}x^j .$ Then $sum_{j=1}^{infty}jx^{j-1}=f'(x)=1/(1-x)^2,$ and $sum_{j=2}^{infty}j(j-1)x^{j-2}=f''(x)=2/(1-x)^3.$
    $endgroup$
    – DanielWainfleet
    1 hour ago












  • $begingroup$
    See : math.stackexchange.com/questions/1711318/…
    $endgroup$
    – lab bhattacharjee
    40 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
2 hours 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
2 hours ago














$begingroup$
If $R>0$ and a power series $sum_{j=0}^{infty}a_jx^j$ converges to $f(x)$ for each $xin (-R,R)$ then $f$ is differentiable on $(-R,R)$ and $sum_{j=1}^{infty}ja_jx^{j-1} =f'(x)$ for all $(-R,R).$ Notice also that the previous sentence now applies with $f$ replaced by $f'$. For example, for $|x|<2$ let $f(x) =1/(1-x)=sum_{j=0}^{infty}x^j .$ Then $sum_{j=1}^{infty}jx^{j-1}=f'(x)=1/(1-x)^2,$ and $sum_{j=2}^{infty}j(j-1)x^{j-2}=f''(x)=2/(1-x)^3.$
$endgroup$
– DanielWainfleet
1 hour ago






$begingroup$
If $R>0$ and a power series $sum_{j=0}^{infty}a_jx^j$ converges to $f(x)$ for each $xin (-R,R)$ then $f$ is differentiable on $(-R,R)$ and $sum_{j=1}^{infty}ja_jx^{j-1} =f'(x)$ for all $(-R,R).$ Notice also that the previous sentence now applies with $f$ replaced by $f'$. For example, for $|x|<2$ let $f(x) =1/(1-x)=sum_{j=0}^{infty}x^j .$ Then $sum_{j=1}^{infty}jx^{j-1}=f'(x)=1/(1-x)^2,$ and $sum_{j=2}^{infty}j(j-1)x^{j-2}=f''(x)=2/(1-x)^3.$
$endgroup$
– DanielWainfleet
1 hour ago














$begingroup$
See : math.stackexchange.com/questions/1711318/…
$endgroup$
– lab bhattacharjee
40 mins ago




$begingroup$
See : math.stackexchange.com/questions/1711318/…
$endgroup$
– lab bhattacharjee
40 mins ago










3 Answers
3






active

oldest

votes


















9












$begingroup$

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
    2 hours ago












  • $begingroup$
    Ok both expressions sum to $e$, i get it, thanks a lot
    $endgroup$
    – user601297
    1 hour 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 $x=1$:
$$sum_{ngeq1}frac{n^2}{n!}=2e$$





Edit



This is a really neat trick that is widely used. Whenever you see an $n^k$ in the numerator, think applying the $xfrac{d}{dx}$ operator $k$ times. Example:
$$e^x-1=sum_{ngeq1}frac{x^n}{n!}$$
Apply $xfrac{d}{dx}$:
$$xe^x=sum_{ngeq1}frac{x}{n!}x^n$$
Apply $xfrac{d}{dx}$:
$$x(x+1)e^x=sum_{ngeq1}frac{n^2}{n!}x^n$$
The pattern continues:
$$left(xfrac{d}{dx}right)^k[e^x-1]=sum_{ngeq1}frac{n^k}{n!}x^n$$





A similar thing can be done with integration. Example:



Evaluate $$S=sum_{ngeq0}frac{(-1)^n}{(2n+2)(2n+1)}$$
Start by recalling that (use geometric series)
$$frac1{1+t^2}=sum_{ngeq0}(-1)^nt^{2n}$$
Then integrate both sides from $0$ to $x$ to get
$$arctan x=sum_{ngeq0}frac{(-1)^n}{2n+1}x^{2n+1}$$
integrate both sides from $0$ to $1$ now to produce
$$S=sum_{ngeq0}frac{(-1)^n}{(2n+2)(2n+1)}=fracpi4-frac12log2$$






share|cite|improve this answer











$endgroup$









  • 1




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










  • $begingroup$
    @user601297 you are very welcome :)
    $endgroup$
    – clathratus
    1 hour ago



















1












$begingroup$

Just to give a slightly different approach,



$$sum_{n=1}^infty{n^2over n!}=sum_{n=1}^infty{nover(n-1)!}=sum_{m=0}^infty{m+1over m!}=sum_{m=0}^infty{mover m!}+e=sum_{m=1}^infty{mover m!}+e=sum_{m=1}^infty{1over(m-1)!}+e=sum_{k=0}^infty{1over k!}+e=e+e$$



The trick in reading this is to note what changes across each equal sign as you proceed from left to right and understand what justifies the equality for each change.






share|cite|improve this answer









$endgroup$




















    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    9












    $begingroup$

    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
      2 hours ago












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
















    9












    $begingroup$

    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
      2 hours ago












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














    9












    9








    9





    $begingroup$

    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 2 hours 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
      2 hours ago












    • $begingroup$
      Ok both expressions sum to $e$, i get it, thanks a lot
      $endgroup$
      – user601297
      1 hour 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
      2 hours ago












    • $begingroup$
      Ok both expressions sum to $e$, i get it, thanks a lot
      $endgroup$
      – user601297
      1 hour 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
    2 hours 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
    2 hours ago














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




    $begingroup$
    Ok both expressions sum to $e$, i get it, thanks a lot
    $endgroup$
    – user601297
    1 hour 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 $x=1$:
    $$sum_{ngeq1}frac{n^2}{n!}=2e$$





    Edit



    This is a really neat trick that is widely used. Whenever you see an $n^k$ in the numerator, think applying the $xfrac{d}{dx}$ operator $k$ times. Example:
    $$e^x-1=sum_{ngeq1}frac{x^n}{n!}$$
    Apply $xfrac{d}{dx}$:
    $$xe^x=sum_{ngeq1}frac{x}{n!}x^n$$
    Apply $xfrac{d}{dx}$:
    $$x(x+1)e^x=sum_{ngeq1}frac{n^2}{n!}x^n$$
    The pattern continues:
    $$left(xfrac{d}{dx}right)^k[e^x-1]=sum_{ngeq1}frac{n^k}{n!}x^n$$





    A similar thing can be done with integration. Example:



    Evaluate $$S=sum_{ngeq0}frac{(-1)^n}{(2n+2)(2n+1)}$$
    Start by recalling that (use geometric series)
    $$frac1{1+t^2}=sum_{ngeq0}(-1)^nt^{2n}$$
    Then integrate both sides from $0$ to $x$ to get
    $$arctan x=sum_{ngeq0}frac{(-1)^n}{2n+1}x^{2n+1}$$
    integrate both sides from $0$ to $1$ now to produce
    $$S=sum_{ngeq0}frac{(-1)^n}{(2n+2)(2n+1)}=fracpi4-frac12log2$$






    share|cite|improve this answer











    $endgroup$









    • 1




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










    • $begingroup$
      @user601297 you are very welcome :)
      $endgroup$
      – clathratus
      1 hour 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 $x=1$:
    $$sum_{ngeq1}frac{n^2}{n!}=2e$$





    Edit



    This is a really neat trick that is widely used. Whenever you see an $n^k$ in the numerator, think applying the $xfrac{d}{dx}$ operator $k$ times. Example:
    $$e^x-1=sum_{ngeq1}frac{x^n}{n!}$$
    Apply $xfrac{d}{dx}$:
    $$xe^x=sum_{ngeq1}frac{x}{n!}x^n$$
    Apply $xfrac{d}{dx}$:
    $$x(x+1)e^x=sum_{ngeq1}frac{n^2}{n!}x^n$$
    The pattern continues:
    $$left(xfrac{d}{dx}right)^k[e^x-1]=sum_{ngeq1}frac{n^k}{n!}x^n$$





    A similar thing can be done with integration. Example:



    Evaluate $$S=sum_{ngeq0}frac{(-1)^n}{(2n+2)(2n+1)}$$
    Start by recalling that (use geometric series)
    $$frac1{1+t^2}=sum_{ngeq0}(-1)^nt^{2n}$$
    Then integrate both sides from $0$ to $x$ to get
    $$arctan x=sum_{ngeq0}frac{(-1)^n}{2n+1}x^{2n+1}$$
    integrate both sides from $0$ to $1$ now to produce
    $$S=sum_{ngeq0}frac{(-1)^n}{(2n+2)(2n+1)}=fracpi4-frac12log2$$






    share|cite|improve this answer











    $endgroup$









    • 1




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










    • $begingroup$
      @user601297 you are very welcome :)
      $endgroup$
      – clathratus
      1 hour 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 $x=1$:
    $$sum_{ngeq1}frac{n^2}{n!}=2e$$





    Edit



    This is a really neat trick that is widely used. Whenever you see an $n^k$ in the numerator, think applying the $xfrac{d}{dx}$ operator $k$ times. Example:
    $$e^x-1=sum_{ngeq1}frac{x^n}{n!}$$
    Apply $xfrac{d}{dx}$:
    $$xe^x=sum_{ngeq1}frac{x}{n!}x^n$$
    Apply $xfrac{d}{dx}$:
    $$x(x+1)e^x=sum_{ngeq1}frac{n^2}{n!}x^n$$
    The pattern continues:
    $$left(xfrac{d}{dx}right)^k[e^x-1]=sum_{ngeq1}frac{n^k}{n!}x^n$$





    A similar thing can be done with integration. Example:



    Evaluate $$S=sum_{ngeq0}frac{(-1)^n}{(2n+2)(2n+1)}$$
    Start by recalling that (use geometric series)
    $$frac1{1+t^2}=sum_{ngeq0}(-1)^nt^{2n}$$
    Then integrate both sides from $0$ to $x$ to get
    $$arctan x=sum_{ngeq0}frac{(-1)^n}{2n+1}x^{2n+1}$$
    integrate both sides from $0$ to $1$ now to produce
    $$S=sum_{ngeq0}frac{(-1)^n}{(2n+2)(2n+1)}=fracpi4-frac12log2$$






    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 $x=1$:
    $$sum_{ngeq1}frac{n^2}{n!}=2e$$





    Edit



    This is a really neat trick that is widely used. Whenever you see an $n^k$ in the numerator, think applying the $xfrac{d}{dx}$ operator $k$ times. Example:
    $$e^x-1=sum_{ngeq1}frac{x^n}{n!}$$
    Apply $xfrac{d}{dx}$:
    $$xe^x=sum_{ngeq1}frac{x}{n!}x^n$$
    Apply $xfrac{d}{dx}$:
    $$x(x+1)e^x=sum_{ngeq1}frac{n^2}{n!}x^n$$
    The pattern continues:
    $$left(xfrac{d}{dx}right)^k[e^x-1]=sum_{ngeq1}frac{n^k}{n!}x^n$$





    A similar thing can be done with integration. Example:



    Evaluate $$S=sum_{ngeq0}frac{(-1)^n}{(2n+2)(2n+1)}$$
    Start by recalling that (use geometric series)
    $$frac1{1+t^2}=sum_{ngeq0}(-1)^nt^{2n}$$
    Then integrate both sides from $0$ to $x$ to get
    $$arctan x=sum_{ngeq0}frac{(-1)^n}{2n+1}x^{2n+1}$$
    integrate both sides from $0$ to $1$ now to produce
    $$S=sum_{ngeq0}frac{(-1)^n}{(2n+2)(2n+1)}=fracpi4-frac12log2$$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited 1 hour ago

























    answered 1 hour ago









    clathratusclathratus

    3,651332




    3,651332








    • 1




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










    • $begingroup$
      @user601297 you are very welcome :)
      $endgroup$
      – clathratus
      1 hour ago














    • 1




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










    • $begingroup$
      @user601297 you are very welcome :)
      $endgroup$
      – clathratus
      1 hour ago








    1




    1




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




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












    $begingroup$
    @user601297 you are very welcome :)
    $endgroup$
    – clathratus
    1 hour ago




    $begingroup$
    @user601297 you are very welcome :)
    $endgroup$
    – clathratus
    1 hour ago











    1












    $begingroup$

    Just to give a slightly different approach,



    $$sum_{n=1}^infty{n^2over n!}=sum_{n=1}^infty{nover(n-1)!}=sum_{m=0}^infty{m+1over m!}=sum_{m=0}^infty{mover m!}+e=sum_{m=1}^infty{mover m!}+e=sum_{m=1}^infty{1over(m-1)!}+e=sum_{k=0}^infty{1over k!}+e=e+e$$



    The trick in reading this is to note what changes across each equal sign as you proceed from left to right and understand what justifies the equality for each change.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Just to give a slightly different approach,



      $$sum_{n=1}^infty{n^2over n!}=sum_{n=1}^infty{nover(n-1)!}=sum_{m=0}^infty{m+1over m!}=sum_{m=0}^infty{mover m!}+e=sum_{m=1}^infty{mover m!}+e=sum_{m=1}^infty{1over(m-1)!}+e=sum_{k=0}^infty{1over k!}+e=e+e$$



      The trick in reading this is to note what changes across each equal sign as you proceed from left to right and understand what justifies the equality for each change.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Just to give a slightly different approach,



        $$sum_{n=1}^infty{n^2over n!}=sum_{n=1}^infty{nover(n-1)!}=sum_{m=0}^infty{m+1over m!}=sum_{m=0}^infty{mover m!}+e=sum_{m=1}^infty{mover m!}+e=sum_{m=1}^infty{1over(m-1)!}+e=sum_{k=0}^infty{1over k!}+e=e+e$$



        The trick in reading this is to note what changes across each equal sign as you proceed from left to right and understand what justifies the equality for each change.






        share|cite|improve this answer









        $endgroup$



        Just to give a slightly different approach,



        $$sum_{n=1}^infty{n^2over n!}=sum_{n=1}^infty{nover(n-1)!}=sum_{m=0}^infty{m+1over m!}=sum_{m=0}^infty{mover m!}+e=sum_{m=1}^infty{mover m!}+e=sum_{m=1}^infty{1over(m-1)!}+e=sum_{k=0}^infty{1over k!}+e=e+e$$



        The trick in reading this is to note what changes across each equal sign as you proceed from left to right and understand what justifies the equality for each change.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 1 hour ago









        Barry CipraBarry Cipra

        59.4k653125




        59.4k653125















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