Calculating an infinite sum: $sumlimits_{n=1}^{infty} 2frac{3^{n-1}n(n-1)}{(n+2)(n+1)!}$












3












$begingroup$


I need to calculate the following formula



$$sum_{n=1}^{infty} 2frac{3^{n-1}n(n-1)}{(n+2)(n+1)!}$$



I know that this sum should be finite, but I don't know how to calculate it. I've done some simplifications and reductions but it get me nowhere. Please help me to calculate it.






calculus sequences-and-series power-series







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Laina Yab is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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    3












    $begingroup$


    I need to calculate the following formula



    $$sum_{n=1}^{infty} 2frac{3^{n-1}n(n-1)}{(n+2)(n+1)!}$$



    I know that this sum should be finite, but I don't know how to calculate it. I've done some simplifications and reductions but it get me nowhere. Please help me to calculate it.






    calculus sequences-and-series power-series







    share|cite|improve this question









    New contributor




    Laina Yab is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.







    $endgroup$















      3












      3








      3


      1



      $begingroup$


      I need to calculate the following formula



      $$sum_{n=1}^{infty} 2frac{3^{n-1}n(n-1)}{(n+2)(n+1)!}$$



      I know that this sum should be finite, but I don't know how to calculate it. I've done some simplifications and reductions but it get me nowhere. Please help me to calculate it.






      calculus sequences-and-series power-series







      share|cite|improve this question









      New contributor




      Laina Yab is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$




      I need to calculate the following formula



      $$sum_{n=1}^{infty} 2frac{3^{n-1}n(n-1)}{(n+2)(n+1)!}$$



      I know that this sum should be finite, but I don't know how to calculate it. I've done some simplifications and reductions but it get me nowhere. Please help me to calculate it.






      calculus sequences-and-series power-series




      calculus sequences-and-series power-series






      share|cite|improve this question









      New contributor




      Laina Yab is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|cite|improve this question









      New contributor




      Laina Yab is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      share|cite|improve this question




      share|cite|improve this question








      edited 1 min ago









      Martin Sleziak

      44.7k9117272




      44.7k9117272






      New contributor




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      asked 1 hour ago









      Laina YabLaina Yab

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      243




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      Laina Yab is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






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          2 Answers
          2






          active

          oldest

          votes


















          4












          $begingroup$

          You may proceed as follows:




          • $e^x = sum_{n=0}^{infty}frac{x^n}{n!} Rightarrow e^x -(1+x+frac{x^2}{2}) = sum_{color{blue}{n=3}}^{infty} frac{x^n}{n!} = sum_{color{blue}{n=1}}^{infty} frac{x^{n+2}}{(n+2)!}$

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


          • $Rightarrow 2xf''(x) = boxed{2sum_{n=1}^{infty} frac{x^{n-1}n(n-1)}{(n+2)!}}$. This is your sum for $x=3$.

          • $Rightarrow 2xf''(x) = 2x frac{e^x(x^2-4x+6)-2(x+3)}{x^4}= 2frac{e^x(x^2-4x+6)-2(x+3)}{x^3}$

          • $x= 3 Rightarrow boxed{2sum_{n=1}^{infty} frac{3^{n-1}n(n-1)}{(n+2)!} =frac{2}{9}(e^3-4)}$






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            I couldn't resist saying this is very elegant
            $endgroup$
            – roman
            39 mins ago



















          3












          $begingroup$

          Hint:



          Write $n(n-1)=(n+2)(n+1)+A(n+2)+B$



          $A=?,B=?$



          Use $sum_{r=0}^inftydfrac{y^r}{r!}=e^y$






          share|cite|improve this answer









          $endgroup$













            Your Answer





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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            4












            $begingroup$

            You may proceed as follows:




            • $e^x = sum_{n=0}^{infty}frac{x^n}{n!} Rightarrow e^x -(1+x+frac{x^2}{2}) = sum_{color{blue}{n=3}}^{infty} frac{x^n}{n!} = sum_{color{blue}{n=1}}^{infty} frac{x^{n+2}}{(n+2)!}$

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


            • $Rightarrow 2xf''(x) = boxed{2sum_{n=1}^{infty} frac{x^{n-1}n(n-1)}{(n+2)!}}$. This is your sum for $x=3$.

            • $Rightarrow 2xf''(x) = 2x frac{e^x(x^2-4x+6)-2(x+3)}{x^4}= 2frac{e^x(x^2-4x+6)-2(x+3)}{x^3}$

            • $x= 3 Rightarrow boxed{2sum_{n=1}^{infty} frac{3^{n-1}n(n-1)}{(n+2)!} =frac{2}{9}(e^3-4)}$






            share|cite|improve this answer









            $endgroup$









            • 1




              $begingroup$
              I couldn't resist saying this is very elegant
              $endgroup$
              – roman
              39 mins ago
















            4












            $begingroup$

            You may proceed as follows:




            • $e^x = sum_{n=0}^{infty}frac{x^n}{n!} Rightarrow e^x -(1+x+frac{x^2}{2}) = sum_{color{blue}{n=3}}^{infty} frac{x^n}{n!} = sum_{color{blue}{n=1}}^{infty} frac{x^{n+2}}{(n+2)!}$

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


            • $Rightarrow 2xf''(x) = boxed{2sum_{n=1}^{infty} frac{x^{n-1}n(n-1)}{(n+2)!}}$. This is your sum for $x=3$.

            • $Rightarrow 2xf''(x) = 2x frac{e^x(x^2-4x+6)-2(x+3)}{x^4}= 2frac{e^x(x^2-4x+6)-2(x+3)}{x^3}$

            • $x= 3 Rightarrow boxed{2sum_{n=1}^{infty} frac{3^{n-1}n(n-1)}{(n+2)!} =frac{2}{9}(e^3-4)}$






            share|cite|improve this answer









            $endgroup$









            • 1




              $begingroup$
              I couldn't resist saying this is very elegant
              $endgroup$
              – roman
              39 mins ago














            4












            4








            4





            $begingroup$

            You may proceed as follows:




            • $e^x = sum_{n=0}^{infty}frac{x^n}{n!} Rightarrow e^x -(1+x+frac{x^2}{2}) = sum_{color{blue}{n=3}}^{infty} frac{x^n}{n!} = sum_{color{blue}{n=1}}^{infty} frac{x^{n+2}}{(n+2)!}$

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


            • $Rightarrow 2xf''(x) = boxed{2sum_{n=1}^{infty} frac{x^{n-1}n(n-1)}{(n+2)!}}$. This is your sum for $x=3$.

            • $Rightarrow 2xf''(x) = 2x frac{e^x(x^2-4x+6)-2(x+3)}{x^4}= 2frac{e^x(x^2-4x+6)-2(x+3)}{x^3}$

            • $x= 3 Rightarrow boxed{2sum_{n=1}^{infty} frac{3^{n-1}n(n-1)}{(n+2)!} =frac{2}{9}(e^3-4)}$






            share|cite|improve this answer









            $endgroup$



            You may proceed as follows:




            • $e^x = sum_{n=0}^{infty}frac{x^n}{n!} Rightarrow e^x -(1+x+frac{x^2}{2}) = sum_{color{blue}{n=3}}^{infty} frac{x^n}{n!} = sum_{color{blue}{n=1}}^{infty} frac{x^{n+2}}{(n+2)!}$

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


            • $Rightarrow 2xf''(x) = boxed{2sum_{n=1}^{infty} frac{x^{n-1}n(n-1)}{(n+2)!}}$. This is your sum for $x=3$.

            • $Rightarrow 2xf''(x) = 2x frac{e^x(x^2-4x+6)-2(x+3)}{x^4}= 2frac{e^x(x^2-4x+6)-2(x+3)}{x^3}$

            • $x= 3 Rightarrow boxed{2sum_{n=1}^{infty} frac{3^{n-1}n(n-1)}{(n+2)!} =frac{2}{9}(e^3-4)}$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 1 hour ago









            trancelocationtrancelocation

            9,7901722




            9,7901722








            • 1




              $begingroup$
              I couldn't resist saying this is very elegant
              $endgroup$
              – roman
              39 mins ago














            • 1




              $begingroup$
              I couldn't resist saying this is very elegant
              $endgroup$
              – roman
              39 mins ago








            1




            1




            $begingroup$
            I couldn't resist saying this is very elegant
            $endgroup$
            – roman
            39 mins ago




            $begingroup$
            I couldn't resist saying this is very elegant
            $endgroup$
            – roman
            39 mins ago











            3












            $begingroup$

            Hint:



            Write $n(n-1)=(n+2)(n+1)+A(n+2)+B$



            $A=?,B=?$



            Use $sum_{r=0}^inftydfrac{y^r}{r!}=e^y$






            share|cite|improve this answer









            $endgroup$


















              3












              $begingroup$

              Hint:



              Write $n(n-1)=(n+2)(n+1)+A(n+2)+B$



              $A=?,B=?$



              Use $sum_{r=0}^inftydfrac{y^r}{r!}=e^y$






              share|cite|improve this answer









              $endgroup$
















                3












                3








                3





                $begingroup$

                Hint:



                Write $n(n-1)=(n+2)(n+1)+A(n+2)+B$



                $A=?,B=?$



                Use $sum_{r=0}^inftydfrac{y^r}{r!}=e^y$






                share|cite|improve this answer









                $endgroup$



                Hint:



                Write $n(n-1)=(n+2)(n+1)+A(n+2)+B$



                $A=?,B=?$



                Use $sum_{r=0}^inftydfrac{y^r}{r!}=e^y$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 1 hour ago









                lab bhattacharjeelab bhattacharjee

                224k15156274




                224k15156274






















                    Laina Yab is a new contributor. Be nice, and check out our Code of Conduct.










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