Proving that $sumlimits_{n=0}^{infty }frac{1}{(2n)!!}=sqrt{e}$












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Proving that $$sum_{n=0}^{infty }frac{1}{(2n)!!}=sqrt{e}$$
Firstly, I tried to check the value with the exponential function at $x=.5$ but I found its terms not equal to the series terms.






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    $begingroup$
    Could you also prove that $quaddisplaystylesum_{n=0}^inftyfrac1{(2n+1)!!}=sqrt e~int_0^1e^{-x^2/2}~dxquad?~$ :-)
    $endgroup$
    – Lucian
    Jan 16 '15 at 19:18


















3












$begingroup$


Proving that $$sum_{n=0}^{infty }frac{1}{(2n)!!}=sqrt{e}$$
Firstly, I tried to check the value with the exponential function at $x=.5$ but I found its terms not equal to the series terms.






sequences-and-series







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Could you also prove that $quaddisplaystylesum_{n=0}^inftyfrac1{(2n+1)!!}=sqrt e~int_0^1e^{-x^2/2}~dxquad?~$ :-)
    $endgroup$
    – Lucian
    Jan 16 '15 at 19:18
















3












3








3


1



$begingroup$


Proving that $$sum_{n=0}^{infty }frac{1}{(2n)!!}=sqrt{e}$$
Firstly, I tried to check the value with the exponential function at $x=.5$ but I found its terms not equal to the series terms.






sequences-and-series







share|cite|improve this question











$endgroup$




Proving that $$sum_{n=0}^{infty }frac{1}{(2n)!!}=sqrt{e}$$
Firstly, I tried to check the value with the exponential function at $x=.5$ but I found its terms not equal to the series terms.






sequences-and-series




sequences-and-series






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share|cite|improve this question













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edited Jan 3 at 20:46









Did

247k23222458




247k23222458










asked Jan 16 '15 at 18:35







user187581















  • 1




    $begingroup$
    Could you also prove that $quaddisplaystylesum_{n=0}^inftyfrac1{(2n+1)!!}=sqrt e~int_0^1e^{-x^2/2}~dxquad?~$ :-)
    $endgroup$
    – Lucian
    Jan 16 '15 at 19:18
















  • 1




    $begingroup$
    Could you also prove that $quaddisplaystylesum_{n=0}^inftyfrac1{(2n+1)!!}=sqrt e~int_0^1e^{-x^2/2}~dxquad?~$ :-)
    $endgroup$
    – Lucian
    Jan 16 '15 at 19:18










1




1




$begingroup$
Could you also prove that $quaddisplaystylesum_{n=0}^inftyfrac1{(2n+1)!!}=sqrt e~int_0^1e^{-x^2/2}~dxquad?~$ :-)
$endgroup$
– Lucian
Jan 16 '15 at 19:18






$begingroup$
Could you also prove that $quaddisplaystylesum_{n=0}^inftyfrac1{(2n+1)!!}=sqrt e~int_0^1e^{-x^2/2}~dxquad?~$ :-)
$endgroup$
– Lucian
Jan 16 '15 at 19:18












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$begingroup$

Note that $$(2n)!! = 2cdot4cdot 6 cdots 2n = 2^n n!$$
so your series is just
$$sum_n frac{(1/2)^n}{n!} = e^{frac{1}{2}}$$






share|cite|improve this answer









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    9












    $begingroup$

    $$e^x = sum_{n=0}^inftydfrac{x^n}{n!}$$



    Plug $x = dfrac{1}{2}$, then $dfrac{x^n}{n!} = dfrac{1}{2^n n!} = dfrac{1}{2 cdot 4 cdot 6 cdots 2n} = dfrac{1}{(2n)!!}$






    share|cite|improve this answer









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      2 Answers
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      2 Answers
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      22












      $begingroup$

      Note that $$(2n)!! = 2cdot4cdot 6 cdots 2n = 2^n n!$$
      so your series is just
      $$sum_n frac{(1/2)^n}{n!} = e^{frac{1}{2}}$$






      share|cite|improve this answer









      $endgroup$


















        22












        $begingroup$

        Note that $$(2n)!! = 2cdot4cdot 6 cdots 2n = 2^n n!$$
        so your series is just
        $$sum_n frac{(1/2)^n}{n!} = e^{frac{1}{2}}$$






        share|cite|improve this answer









        $endgroup$
















          22












          22








          22





          $begingroup$

          Note that $$(2n)!! = 2cdot4cdot 6 cdots 2n = 2^n n!$$
          so your series is just
          $$sum_n frac{(1/2)^n}{n!} = e^{frac{1}{2}}$$






          share|cite|improve this answer









          $endgroup$



          Note that $$(2n)!! = 2cdot4cdot 6 cdots 2n = 2^n n!$$
          so your series is just
          $$sum_n frac{(1/2)^n}{n!} = e^{frac{1}{2}}$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 16 '15 at 18:38









          CrostulCrostul

          27.8k22352




          27.8k22352























              9












              $begingroup$

              $$e^x = sum_{n=0}^inftydfrac{x^n}{n!}$$



              Plug $x = dfrac{1}{2}$, then $dfrac{x^n}{n!} = dfrac{1}{2^n n!} = dfrac{1}{2 cdot 4 cdot 6 cdots 2n} = dfrac{1}{(2n)!!}$






              share|cite|improve this answer









              $endgroup$


















                9












                $begingroup$

                $$e^x = sum_{n=0}^inftydfrac{x^n}{n!}$$



                Plug $x = dfrac{1}{2}$, then $dfrac{x^n}{n!} = dfrac{1}{2^n n!} = dfrac{1}{2 cdot 4 cdot 6 cdots 2n} = dfrac{1}{(2n)!!}$






                share|cite|improve this answer









                $endgroup$
















                  9












                  9








                  9





                  $begingroup$

                  $$e^x = sum_{n=0}^inftydfrac{x^n}{n!}$$



                  Plug $x = dfrac{1}{2}$, then $dfrac{x^n}{n!} = dfrac{1}{2^n n!} = dfrac{1}{2 cdot 4 cdot 6 cdots 2n} = dfrac{1}{(2n)!!}$






                  share|cite|improve this answer









                  $endgroup$



                  $$e^x = sum_{n=0}^inftydfrac{x^n}{n!}$$



                  Plug $x = dfrac{1}{2}$, then $dfrac{x^n}{n!} = dfrac{1}{2^n n!} = dfrac{1}{2 cdot 4 cdot 6 cdots 2n} = dfrac{1}{(2n)!!}$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 16 '15 at 18:38









                  Petite EtincellePetite Etincelle

                  12.4k12148




                  12.4k12148






























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