Finding the sum of a specific alternating series












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$begingroup$


The series $$sum_{n=1}^infty dfrac{(-1)^{n +1} (n+1)}{n!}$$ can be easily proven by using Leibniz's test to be convergent. But I am finding a problem in finding it's sum.
Some help is much required. Thank you






real-analysis calculus







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  • $begingroup$
    Try breaking it into two sums.
    $endgroup$
    – saulspatz
    Jan 16 at 15:10
















1












$begingroup$


The series $$sum_{n=1}^infty dfrac{(-1)^{n +1} (n+1)}{n!}$$ can be easily proven by using Leibniz's test to be convergent. But I am finding a problem in finding it's sum.
Some help is much required. Thank you






real-analysis calculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    Try breaking it into two sums.
    $endgroup$
    – saulspatz
    Jan 16 at 15:10














1












1








1





$begingroup$


The series $$sum_{n=1}^infty dfrac{(-1)^{n +1} (n+1)}{n!}$$ can be easily proven by using Leibniz's test to be convergent. But I am finding a problem in finding it's sum.
Some help is much required. Thank you






real-analysis calculus







share|cite|improve this question











$endgroup$




The series $$sum_{n=1}^infty dfrac{(-1)^{n +1} (n+1)}{n!}$$ can be easily proven by using Leibniz's test to be convergent. But I am finding a problem in finding it's sum.
Some help is much required. Thank you






real-analysis calculus




real-analysis calculus






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share|cite|improve this question













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share|cite|improve this question








edited Jan 16 at 15:09







dame j

















asked Jan 16 at 15:08









dame jdame j

19310




19310












  • $begingroup$
    Try breaking it into two sums.
    $endgroup$
    – saulspatz
    Jan 16 at 15:10


















  • $begingroup$
    Try breaking it into two sums.
    $endgroup$
    – saulspatz
    Jan 16 at 15:10
















$begingroup$
Try breaking it into two sums.
$endgroup$
– saulspatz
Jan 16 at 15:10




$begingroup$
Try breaking it into two sums.
$endgroup$
– saulspatz
Jan 16 at 15:10










4 Answers
4






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3












$begingroup$

$$(-1)^{n+1}cdotdfrac{n+1}{n!}=dfrac{(-1)^{n-1}}{(n-1)!}-dfrac{(-1)^n}{n!}=f(n-1)-f(n)$$



where $f(m)=dfrac{(-1)^m}{m!}$



See https://en.m.wikipedia.org/wiki/Telescoping_series






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    Note that, since all of the relevant sums converge absolutely



    $$sumlimits_{i=1}^infty frac{(-1)^{n+1}(n+1)}{n!} = sumlimits_{i=1}^inftyleft(frac{(-1)^{n+1}}{(n-1)!} + frac{(-1)^{n+1}}{n!}right).$$



    Notice that (again, since everything in sight converges absolutely) the second half of the $n$th term cancels with the first half of the $(n+1)$st term, so our sum telescopes to



    $$frac{(-1)^2}{0!} = 1.$$






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      $
      sum_{n=1}^{infty}frac{(-1)^{n+1}(n+1)}{n!}\
      =sum_{n=0}^{infty}frac{(-1)^{n}(n+2)}{(n+1)!}\
      =sum_{n=0}^{infty}frac{(-1)^{n}}{n!}+sum_{n=0}^{infty}frac{(-1)^{n}}{(n+1)!}\
      =sum_{n=0}^{infty}frac{(-1)^{n}}{n!}+sum_{n=1}^{infty}frac{(-1)^{n+1}}{n!}\
      =sum_{n=0}^{infty}frac{(-1)^{n}}{n!}-sum_{n=1}^{infty}frac{(-1)^n}{n!}\
      =1
      $






      share|cite|improve this answer









      $endgroup$





















        0












        $begingroup$

        Starting with
        $$e^{-x} = sum_{n=0}^{infty} (-1)^n frac{1}{n!} x^n$$
        Multiply by $x$
        $$x e^{-x} = sum_{n=0}^{infty} (-1)^n frac{1}{n!} x^{n+1}$$
        Differentiate with respect to $x$
        $$e^{-x}-x e^{-x} = sum_{n=0}^{infty} (-1)^n frac{n+1}{n!} x^n = 1 - sum_{n=1}^{infty} (-1)^{n+1} frac{n+1}{n!} x^n$$
        Now set $x=1$, with result
        $$0 = 1 - sum_{n=1}^{infty} (-1)^{n+1} frac{n+1}{n!} x^n$$
        so the sum is $1$.






        share|cite|improve this answer









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          4 Answers
          4






          active

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          4 Answers
          4






          active

          oldest

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          active

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          active

          oldest

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          3












          $begingroup$

          $$(-1)^{n+1}cdotdfrac{n+1}{n!}=dfrac{(-1)^{n-1}}{(n-1)!}-dfrac{(-1)^n}{n!}=f(n-1)-f(n)$$



          where $f(m)=dfrac{(-1)^m}{m!}$



          See https://en.m.wikipedia.org/wiki/Telescoping_series






          share|cite|improve this answer









          $endgroup$


















            3












            $begingroup$

            $$(-1)^{n+1}cdotdfrac{n+1}{n!}=dfrac{(-1)^{n-1}}{(n-1)!}-dfrac{(-1)^n}{n!}=f(n-1)-f(n)$$



            where $f(m)=dfrac{(-1)^m}{m!}$



            See https://en.m.wikipedia.org/wiki/Telescoping_series






            share|cite|improve this answer









            $endgroup$
















              3












              3








              3





              $begingroup$

              $$(-1)^{n+1}cdotdfrac{n+1}{n!}=dfrac{(-1)^{n-1}}{(n-1)!}-dfrac{(-1)^n}{n!}=f(n-1)-f(n)$$



              where $f(m)=dfrac{(-1)^m}{m!}$



              See https://en.m.wikipedia.org/wiki/Telescoping_series






              share|cite|improve this answer









              $endgroup$



              $$(-1)^{n+1}cdotdfrac{n+1}{n!}=dfrac{(-1)^{n-1}}{(n-1)!}-dfrac{(-1)^n}{n!}=f(n-1)-f(n)$$



              where $f(m)=dfrac{(-1)^m}{m!}$



              See https://en.m.wikipedia.org/wiki/Telescoping_series







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Jan 16 at 15:12









              lab bhattacharjeelab bhattacharjee

              228k15158279




              228k15158279























                  2












                  $begingroup$

                  Note that, since all of the relevant sums converge absolutely



                  $$sumlimits_{i=1}^infty frac{(-1)^{n+1}(n+1)}{n!} = sumlimits_{i=1}^inftyleft(frac{(-1)^{n+1}}{(n-1)!} + frac{(-1)^{n+1}}{n!}right).$$



                  Notice that (again, since everything in sight converges absolutely) the second half of the $n$th term cancels with the first half of the $(n+1)$st term, so our sum telescopes to



                  $$frac{(-1)^2}{0!} = 1.$$






                  share|cite|improve this answer









                  $endgroup$


















                    2












                    $begingroup$

                    Note that, since all of the relevant sums converge absolutely



                    $$sumlimits_{i=1}^infty frac{(-1)^{n+1}(n+1)}{n!} = sumlimits_{i=1}^inftyleft(frac{(-1)^{n+1}}{(n-1)!} + frac{(-1)^{n+1}}{n!}right).$$



                    Notice that (again, since everything in sight converges absolutely) the second half of the $n$th term cancels with the first half of the $(n+1)$st term, so our sum telescopes to



                    $$frac{(-1)^2}{0!} = 1.$$






                    share|cite|improve this answer









                    $endgroup$
















                      2












                      2








                      2





                      $begingroup$

                      Note that, since all of the relevant sums converge absolutely



                      $$sumlimits_{i=1}^infty frac{(-1)^{n+1}(n+1)}{n!} = sumlimits_{i=1}^inftyleft(frac{(-1)^{n+1}}{(n-1)!} + frac{(-1)^{n+1}}{n!}right).$$



                      Notice that (again, since everything in sight converges absolutely) the second half of the $n$th term cancels with the first half of the $(n+1)$st term, so our sum telescopes to



                      $$frac{(-1)^2}{0!} = 1.$$






                      share|cite|improve this answer









                      $endgroup$



                      Note that, since all of the relevant sums converge absolutely



                      $$sumlimits_{i=1}^infty frac{(-1)^{n+1}(n+1)}{n!} = sumlimits_{i=1}^inftyleft(frac{(-1)^{n+1}}{(n-1)!} + frac{(-1)^{n+1}}{n!}right).$$



                      Notice that (again, since everything in sight converges absolutely) the second half of the $n$th term cancels with the first half of the $(n+1)$st term, so our sum telescopes to



                      $$frac{(-1)^2}{0!} = 1.$$







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Jan 16 at 15:14









                      user3482749user3482749

                      4,3191119




                      4,3191119























                          0












                          $begingroup$

                          $
                          sum_{n=1}^{infty}frac{(-1)^{n+1}(n+1)}{n!}\
                          =sum_{n=0}^{infty}frac{(-1)^{n}(n+2)}{(n+1)!}\
                          =sum_{n=0}^{infty}frac{(-1)^{n}}{n!}+sum_{n=0}^{infty}frac{(-1)^{n}}{(n+1)!}\
                          =sum_{n=0}^{infty}frac{(-1)^{n}}{n!}+sum_{n=1}^{infty}frac{(-1)^{n+1}}{n!}\
                          =sum_{n=0}^{infty}frac{(-1)^{n}}{n!}-sum_{n=1}^{infty}frac{(-1)^n}{n!}\
                          =1
                          $






                          share|cite|improve this answer









                          $endgroup$


















                            0












                            $begingroup$

                            $
                            sum_{n=1}^{infty}frac{(-1)^{n+1}(n+1)}{n!}\
                            =sum_{n=0}^{infty}frac{(-1)^{n}(n+2)}{(n+1)!}\
                            =sum_{n=0}^{infty}frac{(-1)^{n}}{n!}+sum_{n=0}^{infty}frac{(-1)^{n}}{(n+1)!}\
                            =sum_{n=0}^{infty}frac{(-1)^{n}}{n!}+sum_{n=1}^{infty}frac{(-1)^{n+1}}{n!}\
                            =sum_{n=0}^{infty}frac{(-1)^{n}}{n!}-sum_{n=1}^{infty}frac{(-1)^n}{n!}\
                            =1
                            $






                            share|cite|improve this answer









                            $endgroup$
















                              0












                              0








                              0





                              $begingroup$

                              $
                              sum_{n=1}^{infty}frac{(-1)^{n+1}(n+1)}{n!}\
                              =sum_{n=0}^{infty}frac{(-1)^{n}(n+2)}{(n+1)!}\
                              =sum_{n=0}^{infty}frac{(-1)^{n}}{n!}+sum_{n=0}^{infty}frac{(-1)^{n}}{(n+1)!}\
                              =sum_{n=0}^{infty}frac{(-1)^{n}}{n!}+sum_{n=1}^{infty}frac{(-1)^{n+1}}{n!}\
                              =sum_{n=0}^{infty}frac{(-1)^{n}}{n!}-sum_{n=1}^{infty}frac{(-1)^n}{n!}\
                              =1
                              $






                              share|cite|improve this answer









                              $endgroup$



                              $
                              sum_{n=1}^{infty}frac{(-1)^{n+1}(n+1)}{n!}\
                              =sum_{n=0}^{infty}frac{(-1)^{n}(n+2)}{(n+1)!}\
                              =sum_{n=0}^{infty}frac{(-1)^{n}}{n!}+sum_{n=0}^{infty}frac{(-1)^{n}}{(n+1)!}\
                              =sum_{n=0}^{infty}frac{(-1)^{n}}{n!}+sum_{n=1}^{infty}frac{(-1)^{n+1}}{n!}\
                              =sum_{n=0}^{infty}frac{(-1)^{n}}{n!}-sum_{n=1}^{infty}frac{(-1)^n}{n!}\
                              =1
                              $







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Jan 16 at 15:30









                              Fei XuFei Xu

                              12




                              12























                                  0












                                  $begingroup$

                                  Starting with
                                  $$e^{-x} = sum_{n=0}^{infty} (-1)^n frac{1}{n!} x^n$$
                                  Multiply by $x$
                                  $$x e^{-x} = sum_{n=0}^{infty} (-1)^n frac{1}{n!} x^{n+1}$$
                                  Differentiate with respect to $x$
                                  $$e^{-x}-x e^{-x} = sum_{n=0}^{infty} (-1)^n frac{n+1}{n!} x^n = 1 - sum_{n=1}^{infty} (-1)^{n+1} frac{n+1}{n!} x^n$$
                                  Now set $x=1$, with result
                                  $$0 = 1 - sum_{n=1}^{infty} (-1)^{n+1} frac{n+1}{n!} x^n$$
                                  so the sum is $1$.






                                  share|cite|improve this answer









                                  $endgroup$


















                                    0












                                    $begingroup$

                                    Starting with
                                    $$e^{-x} = sum_{n=0}^{infty} (-1)^n frac{1}{n!} x^n$$
                                    Multiply by $x$
                                    $$x e^{-x} = sum_{n=0}^{infty} (-1)^n frac{1}{n!} x^{n+1}$$
                                    Differentiate with respect to $x$
                                    $$e^{-x}-x e^{-x} = sum_{n=0}^{infty} (-1)^n frac{n+1}{n!} x^n = 1 - sum_{n=1}^{infty} (-1)^{n+1} frac{n+1}{n!} x^n$$
                                    Now set $x=1$, with result
                                    $$0 = 1 - sum_{n=1}^{infty} (-1)^{n+1} frac{n+1}{n!} x^n$$
                                    so the sum is $1$.






                                    share|cite|improve this answer









                                    $endgroup$
















                                      0












                                      0








                                      0





                                      $begingroup$

                                      Starting with
                                      $$e^{-x} = sum_{n=0}^{infty} (-1)^n frac{1}{n!} x^n$$
                                      Multiply by $x$
                                      $$x e^{-x} = sum_{n=0}^{infty} (-1)^n frac{1}{n!} x^{n+1}$$
                                      Differentiate with respect to $x$
                                      $$e^{-x}-x e^{-x} = sum_{n=0}^{infty} (-1)^n frac{n+1}{n!} x^n = 1 - sum_{n=1}^{infty} (-1)^{n+1} frac{n+1}{n!} x^n$$
                                      Now set $x=1$, with result
                                      $$0 = 1 - sum_{n=1}^{infty} (-1)^{n+1} frac{n+1}{n!} x^n$$
                                      so the sum is $1$.






                                      share|cite|improve this answer









                                      $endgroup$



                                      Starting with
                                      $$e^{-x} = sum_{n=0}^{infty} (-1)^n frac{1}{n!} x^n$$
                                      Multiply by $x$
                                      $$x e^{-x} = sum_{n=0}^{infty} (-1)^n frac{1}{n!} x^{n+1}$$
                                      Differentiate with respect to $x$
                                      $$e^{-x}-x e^{-x} = sum_{n=0}^{infty} (-1)^n frac{n+1}{n!} x^n = 1 - sum_{n=1}^{infty} (-1)^{n+1} frac{n+1}{n!} x^n$$
                                      Now set $x=1$, with result
                                      $$0 = 1 - sum_{n=1}^{infty} (-1)^{n+1} frac{n+1}{n!} x^n$$
                                      so the sum is $1$.







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Jan 16 at 21:32









                                      awkwardawkward

                                      6,82511026




                                      6,82511026






























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