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
    22 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
    22 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 12 mins ago







user601297

















asked 33 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
    22 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
    22 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
22 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
22 mins ago












3 Answers
3






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
    26 mins ago












  • $begingroup$
    Ok both expressions sum to $e$, i get it, thanks a lot
    $endgroup$
    – user601297
    20 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
    14 mins ago










  • $begingroup$
    @user601297 you are very welcome :)
    $endgroup$
    – clathratus
    12 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









$endgroup$













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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

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    active

    oldest

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    active

    oldest

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    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
      26 mins ago












    • $begingroup$
      Ok both expressions sum to $e$, i get it, thanks a lot
      $endgroup$
      – user601297
      20 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
      26 mins ago












    • $begingroup$
      Ok both expressions sum to $e$, i get it, thanks a lot
      $endgroup$
      – user601297
      20 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 29 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
      26 mins ago












    • $begingroup$
      Ok both expressions sum to $e$, i get it, thanks a lot
      $endgroup$
      – user601297
      20 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
      26 mins ago












    • $begingroup$
      Ok both expressions sum to $e$, i get it, thanks a lot
      $endgroup$
      – user601297
      20 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
    26 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
    26 mins ago














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




    $begingroup$
    Ok both expressions sum to $e$, i get it, thanks a lot
    $endgroup$
    – user601297
    20 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
      14 mins ago










    • $begingroup$
      @user601297 you are very welcome :)
      $endgroup$
      – clathratus
      12 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
      14 mins ago










    • $begingroup$
      @user601297 you are very welcome :)
      $endgroup$
      – clathratus
      12 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 19 mins ago









    clathratusclathratus

    3,651332




    3,651332








    • 1




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










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














    • 1




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










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








    1




    1




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




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












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




    $begingroup$
    @user601297 you are very welcome :)
    $endgroup$
    – clathratus
    12 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









    $endgroup$


















      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









      $endgroup$
















        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









        $endgroup$



        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












        share|cite



        share|cite










        answered 2 mins ago









        Barry CipraBarry Cipra

        59.4k653125




        59.4k653125






























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            list processes belonging to a network namespace

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            5 1 I am on Ubuntu 12.04, and the ip utility does not have ip netns identify <pid> option, I tried installing new iproute , but still, the option identify doesn't seem to be working!. If I were to write a script (or code) to list all processes in a network-namespace, or given a PID, show which network-namespace it belongs to, how should I proceed ? (I need info on a handful of processes, to check if they are in the right netns ) linux process network-namespaces share | improve this question edited Jun 30 '15 at 22:00 Gilles 529k 128 1060 1585 ...
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            list systemd RuntimeDirectory mounts

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            3 I have a systemd service with this declaration: RuntimeDirectory=plex RuntimeDirectoryMode=750 which creates the in memory directory /run/plex . How would I list the capabilities of this mount point as I would do with mount -l ? systemd systemd-mount share | improve this question edited Jan 9 at 17:57 filbranden 7,400 2 8 37 asked Jan 9 at 17:40 adrhc adrhc 247 1 11 ...
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