Finding sum to infinity












4












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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







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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
    38 mins ago


















4












$begingroup$


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
    38 mins ago
















4












4








4


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













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edited 28 mins ago







user601297

















asked 49 mins ago









user601297user601297

37519




37519












  • $begingroup$
    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
    38 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
    38 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
38 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
38 mins ago












3 Answers
3






active

oldest

votes


















8












$begingroup$

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
    42 mins ago












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




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
    30 mins ago










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



















0












$begingroup$

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.






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    3 Answers
    3






    active

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    3 Answers
    3






    active

    oldest

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    active

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    8












    $begingroup$

    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
      42 mins ago












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
















    8












    $begingroup$

    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
      42 mins ago












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














    8












    8








    8





    $begingroup$

    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 45 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
      42 mins ago












    • $begingroup$
      Ok both expressions sum to $e$, i get it, thanks a lot
      $endgroup$
      – user601297
      36 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
      42 mins ago












    • $begingroup$
      Ok both expressions sum to $e$, i get it, thanks a lot
      $endgroup$
      – user601297
      36 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
    42 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
    42 mins ago














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




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




    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
      30 mins ago










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




    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
      30 mins ago










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




    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 3 mins ago

























    answered 34 mins ago









    clathratusclathratus

    3,651332




    3,651332








    • 1




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










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














    • 1




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










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








    1




    1




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




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












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




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











    0












    $begingroup$

    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









    $endgroup$


















      0












      $begingroup$

      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









      $endgroup$
















        0












        0








        0





        $begingroup$

        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 18 mins ago









        Barry CipraBarry Cipra

        59.4k653125




        59.4k653125






























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