How to find sum of the infinite series $frac{n^2}{n!}$ [closed]











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I have tried the Riemann sum also but couldn't find the answer.






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closed as off-topic by amWhy, Shailesh, max_zorn, Hanul Jeon, Brahadeesh Nov 18 at 11:50


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  • Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
    – José Carlos Santos
    Nov 18 at 9:21















up vote
1
down vote

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I have tried the Riemann sum also but couldn't find the answer.






real-analysis sequences-and-series







share|cite|improve this question















closed as off-topic by amWhy, Shailesh, max_zorn, Hanul Jeon, Brahadeesh Nov 18 at 11:50


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Shailesh, max_zorn, Hanul Jeon, Brahadeesh

If this question can be reworded to fit the rules in the help center, please edit the question.













  • Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
    – José Carlos Santos
    Nov 18 at 9:21













up vote
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up vote
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down vote

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2





I have tried the Riemann sum also but couldn't find the answer.






real-analysis sequences-and-series







share|cite|improve this question















I have tried the Riemann sum also but couldn't find the answer.






real-analysis sequences-and-series




real-analysis sequences-and-series






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edited Nov 18 at 9:47









Joey Kilpatrick

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1,183422










asked Nov 18 at 9:18









Ravi Satpute

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closed as off-topic by amWhy, Shailesh, max_zorn, Hanul Jeon, Brahadeesh Nov 18 at 11:50


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Shailesh, max_zorn, Hanul Jeon, Brahadeesh

If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by amWhy, Shailesh, max_zorn, Hanul Jeon, Brahadeesh Nov 18 at 11:50


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Shailesh, max_zorn, Hanul Jeon, Brahadeesh

If this question can be reworded to fit the rules in the help center, please edit the question.












  • Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
    – José Carlos Santos
    Nov 18 at 9:21


















  • Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
    – José Carlos Santos
    Nov 18 at 9:21
















Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Nov 18 at 9:21




Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Nov 18 at 9:21










2 Answers
2






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up vote
1
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accepted










Note that




  • $sum_{n=1}^{infty}frac{n^2}{n!} = sum_{n=1}^{infty}frac{n}{(n-1)!}$

  • $e^x =sum_{n=0}^{infty}frac{x^n}{n!} Rightarrow xe^x = sum_{n=0}^{infty}frac{x^{n+1}}{n!}$

  • $left( xe^x right)' = (1+x)e^x = sum^{infty}_{color{blue}{n=0}}frac{(n+1)x^{n}}{n!} stackrel{i = n+1}{=} sum^{infty}_{color{blue}{i=1}}frac{ix^{i-1}}{(i-1)!}$


$$Rightarrow sum_{n=1}^{infty}frac{n^2}{n!} = sum_{n=1}^{infty}frac{n}{(n-1)!} = (1+1)e = 2e$$






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    up vote
    6
    down vote













    Hint



    Consider
    $$S=sum_{n=0}^infty frac {n^2}{n!}x^n=sum_{n=0}^infty frac {n(n-1)+n}{n!}x^n=sum_{n=0}^infty frac {n(n-1)}{n!}x^n+sum_{n=0}^infty frac {n}{n!}x^n$$ that is to say
    $$S=x^2sum_{n=0}^infty frac {n(n-1)}{n!}x^{n-2}+xsum_{n=0}^infty frac {n}{n!}x^{n-1}$$



    When finished, let $x=1$.






    share|cite|improve this answer




























      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      1
      down vote



      accepted










      Note that




      • $sum_{n=1}^{infty}frac{n^2}{n!} = sum_{n=1}^{infty}frac{n}{(n-1)!}$

      • $e^x =sum_{n=0}^{infty}frac{x^n}{n!} Rightarrow xe^x = sum_{n=0}^{infty}frac{x^{n+1}}{n!}$

      • $left( xe^x right)' = (1+x)e^x = sum^{infty}_{color{blue}{n=0}}frac{(n+1)x^{n}}{n!} stackrel{i = n+1}{=} sum^{infty}_{color{blue}{i=1}}frac{ix^{i-1}}{(i-1)!}$


      $$Rightarrow sum_{n=1}^{infty}frac{n^2}{n!} = sum_{n=1}^{infty}frac{n}{(n-1)!} = (1+1)e = 2e$$






      share|cite|improve this answer

























        up vote
        1
        down vote



        accepted










        Note that




        • $sum_{n=1}^{infty}frac{n^2}{n!} = sum_{n=1}^{infty}frac{n}{(n-1)!}$

        • $e^x =sum_{n=0}^{infty}frac{x^n}{n!} Rightarrow xe^x = sum_{n=0}^{infty}frac{x^{n+1}}{n!}$

        • $left( xe^x right)' = (1+x)e^x = sum^{infty}_{color{blue}{n=0}}frac{(n+1)x^{n}}{n!} stackrel{i = n+1}{=} sum^{infty}_{color{blue}{i=1}}frac{ix^{i-1}}{(i-1)!}$


        $$Rightarrow sum_{n=1}^{infty}frac{n^2}{n!} = sum_{n=1}^{infty}frac{n}{(n-1)!} = (1+1)e = 2e$$






        share|cite|improve this answer























          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          Note that




          • $sum_{n=1}^{infty}frac{n^2}{n!} = sum_{n=1}^{infty}frac{n}{(n-1)!}$

          • $e^x =sum_{n=0}^{infty}frac{x^n}{n!} Rightarrow xe^x = sum_{n=0}^{infty}frac{x^{n+1}}{n!}$

          • $left( xe^x right)' = (1+x)e^x = sum^{infty}_{color{blue}{n=0}}frac{(n+1)x^{n}}{n!} stackrel{i = n+1}{=} sum^{infty}_{color{blue}{i=1}}frac{ix^{i-1}}{(i-1)!}$


          $$Rightarrow sum_{n=1}^{infty}frac{n^2}{n!} = sum_{n=1}^{infty}frac{n}{(n-1)!} = (1+1)e = 2e$$






          share|cite|improve this answer












          Note that




          • $sum_{n=1}^{infty}frac{n^2}{n!} = sum_{n=1}^{infty}frac{n}{(n-1)!}$

          • $e^x =sum_{n=0}^{infty}frac{x^n}{n!} Rightarrow xe^x = sum_{n=0}^{infty}frac{x^{n+1}}{n!}$

          • $left( xe^x right)' = (1+x)e^x = sum^{infty}_{color{blue}{n=0}}frac{(n+1)x^{n}}{n!} stackrel{i = n+1}{=} sum^{infty}_{color{blue}{i=1}}frac{ix^{i-1}}{(i-1)!}$


          $$Rightarrow sum_{n=1}^{infty}frac{n^2}{n!} = sum_{n=1}^{infty}frac{n}{(n-1)!} = (1+1)e = 2e$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 18 at 10:48









          trancelocation

          8,6321520




          8,6321520






















              up vote
              6
              down vote













              Hint



              Consider
              $$S=sum_{n=0}^infty frac {n^2}{n!}x^n=sum_{n=0}^infty frac {n(n-1)+n}{n!}x^n=sum_{n=0}^infty frac {n(n-1)}{n!}x^n+sum_{n=0}^infty frac {n}{n!}x^n$$ that is to say
              $$S=x^2sum_{n=0}^infty frac {n(n-1)}{n!}x^{n-2}+xsum_{n=0}^infty frac {n}{n!}x^{n-1}$$



              When finished, let $x=1$.






              share|cite|improve this answer

























                up vote
                6
                down vote













                Hint



                Consider
                $$S=sum_{n=0}^infty frac {n^2}{n!}x^n=sum_{n=0}^infty frac {n(n-1)+n}{n!}x^n=sum_{n=0}^infty frac {n(n-1)}{n!}x^n+sum_{n=0}^infty frac {n}{n!}x^n$$ that is to say
                $$S=x^2sum_{n=0}^infty frac {n(n-1)}{n!}x^{n-2}+xsum_{n=0}^infty frac {n}{n!}x^{n-1}$$



                When finished, let $x=1$.






                share|cite|improve this answer























                  up vote
                  6
                  down vote










                  up vote
                  6
                  down vote









                  Hint



                  Consider
                  $$S=sum_{n=0}^infty frac {n^2}{n!}x^n=sum_{n=0}^infty frac {n(n-1)+n}{n!}x^n=sum_{n=0}^infty frac {n(n-1)}{n!}x^n+sum_{n=0}^infty frac {n}{n!}x^n$$ that is to say
                  $$S=x^2sum_{n=0}^infty frac {n(n-1)}{n!}x^{n-2}+xsum_{n=0}^infty frac {n}{n!}x^{n-1}$$



                  When finished, let $x=1$.






                  share|cite|improve this answer












                  Hint



                  Consider
                  $$S=sum_{n=0}^infty frac {n^2}{n!}x^n=sum_{n=0}^infty frac {n(n-1)+n}{n!}x^n=sum_{n=0}^infty frac {n(n-1)}{n!}x^n+sum_{n=0}^infty frac {n}{n!}x^n$$ that is to say
                  $$S=x^2sum_{n=0}^infty frac {n(n-1)}{n!}x^{n-2}+xsum_{n=0}^infty frac {n}{n!}x^{n-1}$$



                  When finished, let $x=1$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 18 at 9:32









                  Claude Leibovici

                  117k1156131




                  117k1156131















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