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









    $endgroup$













      Your Answer





      StackExchange.ifUsing("editor", function () {
      return StackExchange.using("mathjaxEditing", function () {
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      });
      });
      }, "mathjax-editing");

      StackExchange.ready(function() {
      var channelOptions = {
      tags: "".split(" "),
      id: "69"
      };
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function() {
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled) {
      StackExchange.using("snippets", function() {
      createEditor();
      });
      }
      else {
      createEditor();
      }
      });

      function createEditor() {
      StackExchange.prepareEditor({
      heartbeatType: 'answer',
      autoActivateHeartbeat: false,
      convertImagesToLinks: true,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      imageUploader: {
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      },
      noCode: true, onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      });


      }
      });














      draft saved

      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3023492%2fsum-n-1-infty-frac-1n-pin12n-2n-1-with-complex-analysis%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      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




                  18.3k31237






























                      draft saved

                      draft discarded




















































                      Thanks for contributing an answer to Mathematics Stack Exchange!


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid …



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      Use MathJax to format equations. MathJax reference.


                      To learn more, see our tips on writing great answers.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3023492%2fsum-n-1-infty-frac-1n-pin12n-2n-1-with-complex-analysis%23new-answer', 'question_page');
                      }
                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      -

                      Popular posts from this blog

                      Probability when a professor distributes a quiz and homework assignment to a class of n students.

                      Image
                      0 0 $begingroup$ Need help with this problem. Suppose our lazy professor collects a quiz and a homework assignment from a class of n students one day, then distributes both the quizzes and the homework assignments back to the class in a random fashion for grading. Each student receives one quiz and one homework assignment to grade. (a) What is the probability that every student receives someone else's quiz to grade, and someone else's homework to grade? (b) What is the probability that no student receives both their own quiz and their own homework assignment to grade? In this case, some students may receive their own quiz, and others may receive their own homework assignment. (c) Compute the limiting probability as n approaches infinity in each case. ...
                      Read more

                      Aardman Animations

                      Image
                      Aardman Animations, Ltd. Aardman Animations Tipo privada Atividade Animação em stop-motion , Animação em CGI Fundação 1972 Fundador(es) Peter Lord David Sproxton Sede Bristol, Inglaterra  Reino Unido Proprietário(s) Dreamworks Pessoas-chave Peter Lord David Sproxton Nick Park Divisões Aardman Features Aardman Digital Aardman Commercials Aardman Broadcast Aardman International Aardman Rights Aardman Effects Aardman 3-D Systems Aardman Nathan Love AardBoiled Website oficial aardman.com ...
                      Read more

                      Are they similar matrix

                      Image
                      0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19 ...
                      Read more