Summing infinite series $sum_n=0^inftyfraccos^2n+1x2n+1$.

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How should one approach questions of this kind:




What is the sum of
$$sum_n=0^inftyfraccos^2n+1x2n+1$$




For $x=0$ this reduces to $1+1/3+1/5+...$ which diverges by the integral test. In fact, the answer is supposed to be $logcot (x/2)$. How do we arrive at it?







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    up vote
    5
    down vote

    favorite
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    How should one approach questions of this kind:




    What is the sum of
    $$sum_n=0^inftyfraccos^2n+1x2n+1$$




    For $x=0$ this reduces to $1+1/3+1/5+...$ which diverges by the integral test. In fact, the answer is supposed to be $logcot (x/2)$. How do we arrive at it?







    share|cite|improve this question
























      up vote
      5
      down vote

      favorite
      3









      up vote
      5
      down vote

      favorite
      3






      3





      How should one approach questions of this kind:




      What is the sum of
      $$sum_n=0^inftyfraccos^2n+1x2n+1$$




      For $x=0$ this reduces to $1+1/3+1/5+...$ which diverges by the integral test. In fact, the answer is supposed to be $logcot (x/2)$. How do we arrive at it?







      share|cite|improve this question














      How should one approach questions of this kind:




      What is the sum of
      $$sum_n=0^inftyfraccos^2n+1x2n+1$$




      For $x=0$ this reduces to $1+1/3+1/5+...$ which diverges by the integral test. In fact, the answer is supposed to be $logcot (x/2)$. How do we arrive at it?









      share|cite|improve this question













      share|cite|improve this question




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      edited Aug 19 at 10:04









      Nosrati

      20.7k41644




      20.7k41644










      asked Aug 19 at 8:46









      Shahab

      4,62612474




      4,62612474




















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          $$sum_n=1^inftyfract^2n+12n+1=tanh^-1t=frac12logfrac1+t1-t.$$
          for $|t|<1$. Substitute $t=cos x$.
          $$sum_n=1^inftyfraccos^2n+1x2n+1=frac12logfrac1+cos x
          1-cos x.$$
          This can be simplified further using half-angle formulae.






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            Where $xneq kpi$ we have $|cos x|<1$ and
            $$sum_n=0^infty cos^2nx=dfrac11-cos^2 x$$
            integration gives
            $$sum_n=0^infty int cos^2nx(-sin x) dx=intdfrac-sin x1-cos^2 x dx$$
            therefore
            $$sum_n=0^infty dfraccos^2n+1x2n+1=lncotdfracx2$$






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              2 Answers
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              up vote
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              $$sum_n=1^inftyfract^2n+12n+1=tanh^-1t=frac12logfrac1+t1-t.$$
              for $|t|<1$. Substitute $t=cos x$.
              $$sum_n=1^inftyfraccos^2n+1x2n+1=frac12logfrac1+cos x
              1-cos x.$$
              This can be simplified further using half-angle formulae.






              share|cite|improve this answer
























                up vote
                8
                down vote













                $$sum_n=1^inftyfract^2n+12n+1=tanh^-1t=frac12logfrac1+t1-t.$$
                for $|t|<1$. Substitute $t=cos x$.
                $$sum_n=1^inftyfraccos^2n+1x2n+1=frac12logfrac1+cos x
                1-cos x.$$
                This can be simplified further using half-angle formulae.






                share|cite|improve this answer






















                  up vote
                  8
                  down vote










                  up vote
                  8
                  down vote









                  $$sum_n=1^inftyfract^2n+12n+1=tanh^-1t=frac12logfrac1+t1-t.$$
                  for $|t|<1$. Substitute $t=cos x$.
                  $$sum_n=1^inftyfraccos^2n+1x2n+1=frac12logfrac1+cos x
                  1-cos x.$$
                  This can be simplified further using half-angle formulae.






                  share|cite|improve this answer












                  $$sum_n=1^inftyfract^2n+12n+1=tanh^-1t=frac12logfrac1+t1-t.$$
                  for $|t|<1$. Substitute $t=cos x$.
                  $$sum_n=1^inftyfraccos^2n+1x2n+1=frac12logfrac1+cos x
                  1-cos x.$$
                  This can be simplified further using half-angle formulae.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Aug 19 at 8:51









                  Lord Shark the Unknown

                  87.5k953114




                  87.5k953114




















                      up vote
                      1
                      down vote













                      Where $xneq kpi$ we have $|cos x|<1$ and
                      $$sum_n=0^infty cos^2nx=dfrac11-cos^2 x$$
                      integration gives
                      $$sum_n=0^infty int cos^2nx(-sin x) dx=intdfrac-sin x1-cos^2 x dx$$
                      therefore
                      $$sum_n=0^infty dfraccos^2n+1x2n+1=lncotdfracx2$$






                      share|cite|improve this answer
























                        up vote
                        1
                        down vote













                        Where $xneq kpi$ we have $|cos x|<1$ and
                        $$sum_n=0^infty cos^2nx=dfrac11-cos^2 x$$
                        integration gives
                        $$sum_n=0^infty int cos^2nx(-sin x) dx=intdfrac-sin x1-cos^2 x dx$$
                        therefore
                        $$sum_n=0^infty dfraccos^2n+1x2n+1=lncotdfracx2$$






                        share|cite|improve this answer






















                          up vote
                          1
                          down vote










                          up vote
                          1
                          down vote









                          Where $xneq kpi$ we have $|cos x|<1$ and
                          $$sum_n=0^infty cos^2nx=dfrac11-cos^2 x$$
                          integration gives
                          $$sum_n=0^infty int cos^2nx(-sin x) dx=intdfrac-sin x1-cos^2 x dx$$
                          therefore
                          $$sum_n=0^infty dfraccos^2n+1x2n+1=lncotdfracx2$$






                          share|cite|improve this answer












                          Where $xneq kpi$ we have $|cos x|<1$ and
                          $$sum_n=0^infty cos^2nx=dfrac11-cos^2 x$$
                          integration gives
                          $$sum_n=0^infty int cos^2nx(-sin x) dx=intdfrac-sin x1-cos^2 x dx$$
                          therefore
                          $$sum_n=0^infty dfraccos^2n+1x2n+1=lncotdfracx2$$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Aug 19 at 10:14









                          Nosrati

                          20.7k41644




                          20.7k41644






















                               

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