$sum_{n=1}^infty frac{(-1)^n pi^{n+1}}{2^n (2n-1)!}$ with complex analysis?












3












$begingroup$


How do I go about evaluating this sum using complex analysis techniques? It is clear that it converges thanks to the ratio test, however I am unsure of how to arrive at the following answer. Thank you.



The answer is $-frac{pi^{3/2} sinbig(sqrt{frac{pi}{2}}big)}{sqrt{2}}$.






sequences-and-series complex-analysis







share|cite|improve this question









$endgroup$








  • 3




    $begingroup$
    Just write the power series for $sin(z)$ and choose $z$ to make it match with your series, up to a constant coefficient. Easier than you might think. Note that $(-1)^npi^n/2^n$ can be written $a^n$.
    $endgroup$
    – Jean-Claude Arbaut
    Dec 3 '18 at 1:47












  • $begingroup$
    Thank you @Jean-ClaudeArbaut. That was a great observation. Do you realize a similar observation with $sum_{n=1}^infty frac{n pi^n}{(2e)^n}$?
    $endgroup$
    – Darkdub
    Dec 3 '18 at 11:38






  • 2




    $begingroup$
    Differentiate $sum_{n=1}^infty x^n$ and note that $pi/(2e)<1$.
    $endgroup$
    – Jean-Claude Arbaut
    Dec 3 '18 at 11:38


















3












$begingroup$


How do I go about evaluating this sum using complex analysis techniques? It is clear that it converges thanks to the ratio test, however I am unsure of how to arrive at the following answer. Thank you.



The answer is $-frac{pi^{3/2} sinbig(sqrt{frac{pi}{2}}big)}{sqrt{2}}$.






sequences-and-series complex-analysis







share|cite|improve this question









$endgroup$








  • 3




    $begingroup$
    Just write the power series for $sin(z)$ and choose $z$ to make it match with your series, up to a constant coefficient. Easier than you might think. Note that $(-1)^npi^n/2^n$ can be written $a^n$.
    $endgroup$
    – Jean-Claude Arbaut
    Dec 3 '18 at 1:47












  • $begingroup$
    Thank you @Jean-ClaudeArbaut. That was a great observation. Do you realize a similar observation with $sum_{n=1}^infty frac{n pi^n}{(2e)^n}$?
    $endgroup$
    – Darkdub
    Dec 3 '18 at 11:38






  • 2




    $begingroup$
    Differentiate $sum_{n=1}^infty x^n$ and note that $pi/(2e)<1$.
    $endgroup$
    – Jean-Claude Arbaut
    Dec 3 '18 at 11:38
















3












3








3





$begingroup$


How do I go about evaluating this sum using complex analysis techniques? It is clear that it converges thanks to the ratio test, however I am unsure of how to arrive at the following answer. Thank you.



The answer is $-frac{pi^{3/2} sinbig(sqrt{frac{pi}{2}}big)}{sqrt{2}}$.






sequences-and-series complex-analysis







share|cite|improve this question









$endgroup$




How do I go about evaluating this sum using complex analysis techniques? It is clear that it converges thanks to the ratio test, however I am unsure of how to arrive at the following answer. Thank you.



The answer is $-frac{pi^{3/2} sinbig(sqrt{frac{pi}{2}}big)}{sqrt{2}}$.






sequences-and-series complex-analysis




sequences-and-series complex-analysis






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 3 '18 at 1:45









DarkdubDarkdub

9816




9816








  • 3




    $begingroup$
    Just write the power series for $sin(z)$ and choose $z$ to make it match with your series, up to a constant coefficient. Easier than you might think. Note that $(-1)^npi^n/2^n$ can be written $a^n$.
    $endgroup$
    – Jean-Claude Arbaut
    Dec 3 '18 at 1:47












  • $begingroup$
    Thank you @Jean-ClaudeArbaut. That was a great observation. Do you realize a similar observation with $sum_{n=1}^infty frac{n pi^n}{(2e)^n}$?
    $endgroup$
    – Darkdub
    Dec 3 '18 at 11:38






  • 2




    $begingroup$
    Differentiate $sum_{n=1}^infty x^n$ and note that $pi/(2e)<1$.
    $endgroup$
    – Jean-Claude Arbaut
    Dec 3 '18 at 11:38
















  • 3




    $begingroup$
    Just write the power series for $sin(z)$ and choose $z$ to make it match with your series, up to a constant coefficient. Easier than you might think. Note that $(-1)^npi^n/2^n$ can be written $a^n$.
    $endgroup$
    – Jean-Claude Arbaut
    Dec 3 '18 at 1:47












  • $begingroup$
    Thank you @Jean-ClaudeArbaut. That was a great observation. Do you realize a similar observation with $sum_{n=1}^infty frac{n pi^n}{(2e)^n}$?
    $endgroup$
    – Darkdub
    Dec 3 '18 at 11:38






  • 2




    $begingroup$
    Differentiate $sum_{n=1}^infty x^n$ and note that $pi/(2e)<1$.
    $endgroup$
    – Jean-Claude Arbaut
    Dec 3 '18 at 11:38










3




3




$begingroup$
Just write the power series for $sin(z)$ and choose $z$ to make it match with your series, up to a constant coefficient. Easier than you might think. Note that $(-1)^npi^n/2^n$ can be written $a^n$.
$endgroup$
– Jean-Claude Arbaut
Dec 3 '18 at 1:47






$begingroup$
Just write the power series for $sin(z)$ and choose $z$ to make it match with your series, up to a constant coefficient. Easier than you might think. Note that $(-1)^npi^n/2^n$ can be written $a^n$.
$endgroup$
– Jean-Claude Arbaut
Dec 3 '18 at 1:47














$begingroup$
Thank you @Jean-ClaudeArbaut. That was a great observation. Do you realize a similar observation with $sum_{n=1}^infty frac{n pi^n}{(2e)^n}$?
$endgroup$
– Darkdub
Dec 3 '18 at 11:38




$begingroup$
Thank you @Jean-ClaudeArbaut. That was a great observation. Do you realize a similar observation with $sum_{n=1}^infty frac{n pi^n}{(2e)^n}$?
$endgroup$
– Darkdub
Dec 3 '18 at 11:38




2




2




$begingroup$
Differentiate $sum_{n=1}^infty x^n$ and note that $pi/(2e)<1$.
$endgroup$
– Jean-Claude Arbaut
Dec 3 '18 at 11:38






$begingroup$
Differentiate $sum_{n=1}^infty x^n$ and note that $pi/(2e)<1$.
$endgroup$
– Jean-Claude Arbaut
Dec 3 '18 at 11:38












3 Answers
3






active

oldest

votes


















2












$begingroup$

Someone has already posted the same answer, but I spent 10 minutes on this, and I'm not about to just delete it.



First of all we recall that, for all $x$,
$$sin x=sum_{ngeq0}frac{(-1)^nx^{2n+1}}{(2n+1)!}$$
Hence we have that
$$
begin{align}
frac{-pi^{3/2}}{2^{1/2}}sinsqrt{fracpi2}=&-frac{pi^{3/2}}{2^{1/2}}sum_{ngeq0}frac{(-1)^n}{(2n+1)!}bigg(frac{pi^{1/2}}{2^{1/2}}bigg)^{2n+1}\
=&frac{pi^{3/2}}{2^{1/2}}sum_{ngeq0}frac{(-1)^{n+1}}{(2n+1)!}frac{pi^{frac{2n+1}2}}{2^{frac{2n+1}2}}\
=&sum_{ngeq0}frac{(-1)^{n+1}}{(2n+1)!}frac{pi^{frac{2n+4}2}}{2^{frac{2n+2}2}}\
=&sum_{ngeq0}frac{(-1)^{n+1}}{(2n+1)!}frac{pi^{n+2}}{2^{n+1}}\
=&sum_{mgeq1}frac{(-1)^m}{(2m-1)!}frac{pi^{m+1}}{2^m}
end{align}
$$
This last step coming from the change of index $n=m-1$






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    G Cab posted their answer just when I was about to post mine, and so I made a similar decision. :)
    $endgroup$
    – Matt A Pelto
    Dec 3 '18 at 2:34






  • 2




    $begingroup$
    @MattAPelto Great minds think alike ;)
    $endgroup$
    – clathratus
    Dec 3 '18 at 2:35



















3












$begingroup$

The Taylor series for $sin(z)$ centered at $0$ is $sum_{n=0}^infty frac{(-1)^n }{ (2n+1)!}z^{2n+1}$.



So for the given series, we have
begin{aligned}sum_{n=1}^infty frac{(-1)^n pi^{n+1}}{2^n (2n-1)!}&=sum_{n=0}^infty frac{(-1)^{n+1} pi^{n+2}}{2^{n+1} (2(n+1)-1)!}
\&= -sqrt{frac{pi^3}{2}}sum_{n=0}^infty frac{(-1)^n }{(2n+1)!}left(sqrt{frac{pi}{2}}right)^{2n+1}
\&=-sqrt{frac{pi^3}{2}}sinleft(sqrt{frac{pi}{2}}right)end{aligned}






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    $$
    eqalign{
    & sumlimits_{1, le ,n} {{{left( { - 1} right)^n pi ^{,n + 1} } over {2^{,n} left( {2n - 1} right)!}}}
    = sumlimits_{0, le ,n} {{{left( { - 1} right)^{n + 1} pi ^{,n + 2} } over {2^{,n + 1} left( {2n + 1} right)!}}} = cr
    & = pi sumlimits_{0, le ,n} {{{i^{,2n + 2} sqrt {pi /2} ^{,2n + 2} } over {left( {2n + 1} right)!}}}
    = i,pi sqrt {pi /2} sumlimits_{0, le ,n} {{{i^{,2n + 1} sqrt {pi /2} ^{,2n + 1} } over {left( {2n + 1} right)!}}} = cr
    & = i,pi sqrt {pi /2} sinh left( {isqrt {pi /2} } right) = - ,pi sqrt {pi /2} sin left( {sqrt {pi /2} } right) cr}
    $$






    share|cite|improve this answer









    $endgroup$













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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      Someone has already posted the same answer, but I spent 10 minutes on this, and I'm not about to just delete it.



      First of all we recall that, for all $x$,
      $$sin x=sum_{ngeq0}frac{(-1)^nx^{2n+1}}{(2n+1)!}$$
      Hence we have that
      $$
      begin{align}
      frac{-pi^{3/2}}{2^{1/2}}sinsqrt{fracpi2}=&-frac{pi^{3/2}}{2^{1/2}}sum_{ngeq0}frac{(-1)^n}{(2n+1)!}bigg(frac{pi^{1/2}}{2^{1/2}}bigg)^{2n+1}\
      =&frac{pi^{3/2}}{2^{1/2}}sum_{ngeq0}frac{(-1)^{n+1}}{(2n+1)!}frac{pi^{frac{2n+1}2}}{2^{frac{2n+1}2}}\
      =&sum_{ngeq0}frac{(-1)^{n+1}}{(2n+1)!}frac{pi^{frac{2n+4}2}}{2^{frac{2n+2}2}}\
      =&sum_{ngeq0}frac{(-1)^{n+1}}{(2n+1)!}frac{pi^{n+2}}{2^{n+1}}\
      =&sum_{mgeq1}frac{(-1)^m}{(2m-1)!}frac{pi^{m+1}}{2^m}
      end{align}
      $$
      This last step coming from the change of index $n=m-1$






      share|cite|improve this answer









      $endgroup$









      • 1




        $begingroup$
        G Cab posted their answer just when I was about to post mine, and so I made a similar decision. :)
        $endgroup$
        – Matt A Pelto
        Dec 3 '18 at 2:34






      • 2




        $begingroup$
        @MattAPelto Great minds think alike ;)
        $endgroup$
        – clathratus
        Dec 3 '18 at 2:35
















      2












      $begingroup$

      Someone has already posted the same answer, but I spent 10 minutes on this, and I'm not about to just delete it.



      First of all we recall that, for all $x$,
      $$sin x=sum_{ngeq0}frac{(-1)^nx^{2n+1}}{(2n+1)!}$$
      Hence we have that
      $$
      begin{align}
      frac{-pi^{3/2}}{2^{1/2}}sinsqrt{fracpi2}=&-frac{pi^{3/2}}{2^{1/2}}sum_{ngeq0}frac{(-1)^n}{(2n+1)!}bigg(frac{pi^{1/2}}{2^{1/2}}bigg)^{2n+1}\
      =&frac{pi^{3/2}}{2^{1/2}}sum_{ngeq0}frac{(-1)^{n+1}}{(2n+1)!}frac{pi^{frac{2n+1}2}}{2^{frac{2n+1}2}}\
      =&sum_{ngeq0}frac{(-1)^{n+1}}{(2n+1)!}frac{pi^{frac{2n+4}2}}{2^{frac{2n+2}2}}\
      =&sum_{ngeq0}frac{(-1)^{n+1}}{(2n+1)!}frac{pi^{n+2}}{2^{n+1}}\
      =&sum_{mgeq1}frac{(-1)^m}{(2m-1)!}frac{pi^{m+1}}{2^m}
      end{align}
      $$
      This last step coming from the change of index $n=m-1$






      share|cite|improve this answer









      $endgroup$









      • 1




        $begingroup$
        G Cab posted their answer just when I was about to post mine, and so I made a similar decision. :)
        $endgroup$
        – Matt A Pelto
        Dec 3 '18 at 2:34






      • 2




        $begingroup$
        @MattAPelto Great minds think alike ;)
        $endgroup$
        – clathratus
        Dec 3 '18 at 2:35














      2












      2








      2





      $begingroup$

      Someone has already posted the same answer, but I spent 10 minutes on this, and I'm not about to just delete it.



      First of all we recall that, for all $x$,
      $$sin x=sum_{ngeq0}frac{(-1)^nx^{2n+1}}{(2n+1)!}$$
      Hence we have that
      $$
      begin{align}
      frac{-pi^{3/2}}{2^{1/2}}sinsqrt{fracpi2}=&-frac{pi^{3/2}}{2^{1/2}}sum_{ngeq0}frac{(-1)^n}{(2n+1)!}bigg(frac{pi^{1/2}}{2^{1/2}}bigg)^{2n+1}\
      =&frac{pi^{3/2}}{2^{1/2}}sum_{ngeq0}frac{(-1)^{n+1}}{(2n+1)!}frac{pi^{frac{2n+1}2}}{2^{frac{2n+1}2}}\
      =&sum_{ngeq0}frac{(-1)^{n+1}}{(2n+1)!}frac{pi^{frac{2n+4}2}}{2^{frac{2n+2}2}}\
      =&sum_{ngeq0}frac{(-1)^{n+1}}{(2n+1)!}frac{pi^{n+2}}{2^{n+1}}\
      =&sum_{mgeq1}frac{(-1)^m}{(2m-1)!}frac{pi^{m+1}}{2^m}
      end{align}
      $$
      This last step coming from the change of index $n=m-1$






      share|cite|improve this answer









      $endgroup$



      Someone has already posted the same answer, but I spent 10 minutes on this, and I'm not about to just delete it.



      First of all we recall that, for all $x$,
      $$sin x=sum_{ngeq0}frac{(-1)^nx^{2n+1}}{(2n+1)!}$$
      Hence we have that
      $$
      begin{align}
      frac{-pi^{3/2}}{2^{1/2}}sinsqrt{fracpi2}=&-frac{pi^{3/2}}{2^{1/2}}sum_{ngeq0}frac{(-1)^n}{(2n+1)!}bigg(frac{pi^{1/2}}{2^{1/2}}bigg)^{2n+1}\
      =&frac{pi^{3/2}}{2^{1/2}}sum_{ngeq0}frac{(-1)^{n+1}}{(2n+1)!}frac{pi^{frac{2n+1}2}}{2^{frac{2n+1}2}}\
      =&sum_{ngeq0}frac{(-1)^{n+1}}{(2n+1)!}frac{pi^{frac{2n+4}2}}{2^{frac{2n+2}2}}\
      =&sum_{ngeq0}frac{(-1)^{n+1}}{(2n+1)!}frac{pi^{n+2}}{2^{n+1}}\
      =&sum_{mgeq1}frac{(-1)^m}{(2m-1)!}frac{pi^{m+1}}{2^m}
      end{align}
      $$
      This last step coming from the change of index $n=m-1$







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Dec 3 '18 at 2:28









      clathratusclathratus

      3,551332




      3,551332








      • 1




        $begingroup$
        G Cab posted their answer just when I was about to post mine, and so I made a similar decision. :)
        $endgroup$
        – Matt A Pelto
        Dec 3 '18 at 2:34






      • 2




        $begingroup$
        @MattAPelto Great minds think alike ;)
        $endgroup$
        – clathratus
        Dec 3 '18 at 2:35














      • 1




        $begingroup$
        G Cab posted their answer just when I was about to post mine, and so I made a similar decision. :)
        $endgroup$
        – Matt A Pelto
        Dec 3 '18 at 2:34






      • 2




        $begingroup$
        @MattAPelto Great minds think alike ;)
        $endgroup$
        – clathratus
        Dec 3 '18 at 2:35








      1




      1




      $begingroup$
      G Cab posted their answer just when I was about to post mine, and so I made a similar decision. :)
      $endgroup$
      – Matt A Pelto
      Dec 3 '18 at 2:34




      $begingroup$
      G Cab posted their answer just when I was about to post mine, and so I made a similar decision. :)
      $endgroup$
      – Matt A Pelto
      Dec 3 '18 at 2:34




      2




      2




      $begingroup$
      @MattAPelto Great minds think alike ;)
      $endgroup$
      – clathratus
      Dec 3 '18 at 2:35




      $begingroup$
      @MattAPelto Great minds think alike ;)
      $endgroup$
      – clathratus
      Dec 3 '18 at 2:35











      3












      $begingroup$

      The Taylor series for $sin(z)$ centered at $0$ is $sum_{n=0}^infty frac{(-1)^n }{ (2n+1)!}z^{2n+1}$.



      So for the given series, we have
      begin{aligned}sum_{n=1}^infty frac{(-1)^n pi^{n+1}}{2^n (2n-1)!}&=sum_{n=0}^infty frac{(-1)^{n+1} pi^{n+2}}{2^{n+1} (2(n+1)-1)!}
      \&= -sqrt{frac{pi^3}{2}}sum_{n=0}^infty frac{(-1)^n }{(2n+1)!}left(sqrt{frac{pi}{2}}right)^{2n+1}
      \&=-sqrt{frac{pi^3}{2}}sinleft(sqrt{frac{pi}{2}}right)end{aligned}






      share|cite|improve this answer









      $endgroup$


















        3












        $begingroup$

        The Taylor series for $sin(z)$ centered at $0$ is $sum_{n=0}^infty frac{(-1)^n }{ (2n+1)!}z^{2n+1}$.



        So for the given series, we have
        begin{aligned}sum_{n=1}^infty frac{(-1)^n pi^{n+1}}{2^n (2n-1)!}&=sum_{n=0}^infty frac{(-1)^{n+1} pi^{n+2}}{2^{n+1} (2(n+1)-1)!}
        \&= -sqrt{frac{pi^3}{2}}sum_{n=0}^infty frac{(-1)^n }{(2n+1)!}left(sqrt{frac{pi}{2}}right)^{2n+1}
        \&=-sqrt{frac{pi^3}{2}}sinleft(sqrt{frac{pi}{2}}right)end{aligned}






        share|cite|improve this answer









        $endgroup$
















          3












          3








          3





          $begingroup$

          The Taylor series for $sin(z)$ centered at $0$ is $sum_{n=0}^infty frac{(-1)^n }{ (2n+1)!}z^{2n+1}$.



          So for the given series, we have
          begin{aligned}sum_{n=1}^infty frac{(-1)^n pi^{n+1}}{2^n (2n-1)!}&=sum_{n=0}^infty frac{(-1)^{n+1} pi^{n+2}}{2^{n+1} (2(n+1)-1)!}
          \&= -sqrt{frac{pi^3}{2}}sum_{n=0}^infty frac{(-1)^n }{(2n+1)!}left(sqrt{frac{pi}{2}}right)^{2n+1}
          \&=-sqrt{frac{pi^3}{2}}sinleft(sqrt{frac{pi}{2}}right)end{aligned}






          share|cite|improve this answer









          $endgroup$



          The Taylor series for $sin(z)$ centered at $0$ is $sum_{n=0}^infty frac{(-1)^n }{ (2n+1)!}z^{2n+1}$.



          So for the given series, we have
          begin{aligned}sum_{n=1}^infty frac{(-1)^n pi^{n+1}}{2^n (2n-1)!}&=sum_{n=0}^infty frac{(-1)^{n+1} pi^{n+2}}{2^{n+1} (2(n+1)-1)!}
          \&= -sqrt{frac{pi^3}{2}}sum_{n=0}^infty frac{(-1)^n }{(2n+1)!}left(sqrt{frac{pi}{2}}right)^{2n+1}
          \&=-sqrt{frac{pi^3}{2}}sinleft(sqrt{frac{pi}{2}}right)end{aligned}







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 3 '18 at 2:21









          Matt A PeltoMatt A Pelto

          2,477620




          2,477620























              1












              $begingroup$

              $$
              eqalign{
              & sumlimits_{1, le ,n} {{{left( { - 1} right)^n pi ^{,n + 1} } over {2^{,n} left( {2n - 1} right)!}}}
              = sumlimits_{0, le ,n} {{{left( { - 1} right)^{n + 1} pi ^{,n + 2} } over {2^{,n + 1} left( {2n + 1} right)!}}} = cr
              & = pi sumlimits_{0, le ,n} {{{i^{,2n + 2} sqrt {pi /2} ^{,2n + 2} } over {left( {2n + 1} right)!}}}
              = i,pi sqrt {pi /2} sumlimits_{0, le ,n} {{{i^{,2n + 1} sqrt {pi /2} ^{,2n + 1} } over {left( {2n + 1} right)!}}} = cr
              & = i,pi sqrt {pi /2} sinh left( {isqrt {pi /2} } right) = - ,pi sqrt {pi /2} sin left( {sqrt {pi /2} } right) cr}
              $$






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                $$
                eqalign{
                & sumlimits_{1, le ,n} {{{left( { - 1} right)^n pi ^{,n + 1} } over {2^{,n} left( {2n - 1} right)!}}}
                = sumlimits_{0, le ,n} {{{left( { - 1} right)^{n + 1} pi ^{,n + 2} } over {2^{,n + 1} left( {2n + 1} right)!}}} = cr
                & = pi sumlimits_{0, le ,n} {{{i^{,2n + 2} sqrt {pi /2} ^{,2n + 2} } over {left( {2n + 1} right)!}}}
                = i,pi sqrt {pi /2} sumlimits_{0, le ,n} {{{i^{,2n + 1} sqrt {pi /2} ^{,2n + 1} } over {left( {2n + 1} right)!}}} = cr
                & = i,pi sqrt {pi /2} sinh left( {isqrt {pi /2} } right) = - ,pi sqrt {pi /2} sin left( {sqrt {pi /2} } right) cr}
                $$






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  $$
                  eqalign{
                  & sumlimits_{1, le ,n} {{{left( { - 1} right)^n pi ^{,n + 1} } over {2^{,n} left( {2n - 1} right)!}}}
                  = sumlimits_{0, le ,n} {{{left( { - 1} right)^{n + 1} pi ^{,n + 2} } over {2^{,n + 1} left( {2n + 1} right)!}}} = cr
                  & = pi sumlimits_{0, le ,n} {{{i^{,2n + 2} sqrt {pi /2} ^{,2n + 2} } over {left( {2n + 1} right)!}}}
                  = i,pi sqrt {pi /2} sumlimits_{0, le ,n} {{{i^{,2n + 1} sqrt {pi /2} ^{,2n + 1} } over {left( {2n + 1} right)!}}} = cr
                  & = i,pi sqrt {pi /2} sinh left( {isqrt {pi /2} } right) = - ,pi sqrt {pi /2} sin left( {sqrt {pi /2} } right) cr}
                  $$






                  share|cite|improve this answer









                  $endgroup$



                  $$
                  eqalign{
                  & sumlimits_{1, le ,n} {{{left( { - 1} right)^n pi ^{,n + 1} } over {2^{,n} left( {2n - 1} right)!}}}
                  = sumlimits_{0, le ,n} {{{left( { - 1} right)^{n + 1} pi ^{,n + 2} } over {2^{,n + 1} left( {2n + 1} right)!}}} = cr
                  & = pi sumlimits_{0, le ,n} {{{i^{,2n + 2} sqrt {pi /2} ^{,2n + 2} } over {left( {2n + 1} right)!}}}
                  = i,pi sqrt {pi /2} sumlimits_{0, le ,n} {{{i^{,2n + 1} sqrt {pi /2} ^{,2n + 1} } over {left( {2n + 1} right)!}}} = cr
                  & = i,pi sqrt {pi /2} sinh left( {isqrt {pi /2} } right) = - ,pi sqrt {pi /2} sin left( {sqrt {pi /2} } right) cr}
                  $$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 3 '18 at 2:20









                  G CabG Cab

                  18.3k31237




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