Sum involving $ln{(2)}$












5












$begingroup$


I got this sum. How do can this sum be equal to $8ln{(2)}?$



$$sum_{n=2}^{infty}frac{(-1)^n}{n}left[frac{35n-37}{(2n-1)(n-1)^2}+frac{35n+37}{(2n+1)(n+1)^2}right]=8ln{(2)}$$



I have try to expand out the sum but it is too messy. Dealing the sum in this form, I haven't got any idea.



Any help.






real-analysis sequences-and-series







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$endgroup$

















    5












    $begingroup$


    I got this sum. How do can this sum be equal to $8ln{(2)}?$



    $$sum_{n=2}^{infty}frac{(-1)^n}{n}left[frac{35n-37}{(2n-1)(n-1)^2}+frac{35n+37}{(2n+1)(n+1)^2}right]=8ln{(2)}$$



    I have try to expand out the sum but it is too messy. Dealing the sum in this form, I haven't got any idea.



    Any help.






    real-analysis sequences-and-series







    share|cite|improve this question











    $endgroup$















      5












      5








      5


      4



      $begingroup$


      I got this sum. How do can this sum be equal to $8ln{(2)}?$



      $$sum_{n=2}^{infty}frac{(-1)^n}{n}left[frac{35n-37}{(2n-1)(n-1)^2}+frac{35n+37}{(2n+1)(n+1)^2}right]=8ln{(2)}$$



      I have try to expand out the sum but it is too messy. Dealing the sum in this form, I haven't got any idea.



      Any help.






      real-analysis sequences-and-series







      share|cite|improve this question











      $endgroup$




      I got this sum. How do can this sum be equal to $8ln{(2)}?$



      $$sum_{n=2}^{infty}frac{(-1)^n}{n}left[frac{35n-37}{(2n-1)(n-1)^2}+frac{35n+37}{(2n+1)(n+1)^2}right]=8ln{(2)}$$



      I have try to expand out the sum but it is too messy. Dealing the sum in this form, I haven't got any idea.



      Any help.






      real-analysis sequences-and-series




      real-analysis sequences-and-series






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 31 at 16:42







      Dragon

















      asked Jan 31 at 13:20









      DragonDragon

      655




      655






















          1 Answer
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          6












          $begingroup$

          First note that
          begin{eqnarray*}
          &&frac{1}{n}left[frac{35n-37}{(2n-1)(n-1)^2}+frac{35n+37}{(2n+1)(n+1)^2}right]\
          &=&frac{74}{n}+frac2{(n+1)^2}-frac2{(n-1)^2}+frac{41}{n+1}+frac{41}{n-1}-frac{156}{2n+1}-frac{156}{2n-1}
          end{eqnarray*}
          and
          begin{eqnarray*}
          &&sum_{n=2}^{infty}frac{(-1)^n}{n-1}=ln2,sum_{n=2}^{infty}frac{(-1)^n}{n}=1-ln2,sum_{n=2}^{infty}frac{(-1)^n}{n+1}=-frac12+ln2,\
          &&sum_{n=2}^{infty}frac{(-1)^n}{2n-1}=frac{4-pi}{4},sum_{n=2}^{infty}frac{(-1)^n}{2n+1}=frac{-8+3pi}{12},\
          &&sum_{n=2}^{infty}(-1)^nbigg[frac2{(n+1)^2}-frac2{(n-1)^2}bigg]\
          &=&2sum_{n=1}^{infty}bigg[frac1{(2n+1)^2}-frac1{(2n-1)^2}bigg]-2sum_{n=2}^{infty}bigg[frac1{(2n)^2}-frac1{(2n-2)^2}bigg]\
          &=&-2+frac12=-frac32.
          end{eqnarray*}
          Then you can put them to give the answer.






          share|cite|improve this answer











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            1 Answer
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            active

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            active

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            6












            $begingroup$

            First note that
            begin{eqnarray*}
            &&frac{1}{n}left[frac{35n-37}{(2n-1)(n-1)^2}+frac{35n+37}{(2n+1)(n+1)^2}right]\
            &=&frac{74}{n}+frac2{(n+1)^2}-frac2{(n-1)^2}+frac{41}{n+1}+frac{41}{n-1}-frac{156}{2n+1}-frac{156}{2n-1}
            end{eqnarray*}
            and
            begin{eqnarray*}
            &&sum_{n=2}^{infty}frac{(-1)^n}{n-1}=ln2,sum_{n=2}^{infty}frac{(-1)^n}{n}=1-ln2,sum_{n=2}^{infty}frac{(-1)^n}{n+1}=-frac12+ln2,\
            &&sum_{n=2}^{infty}frac{(-1)^n}{2n-1}=frac{4-pi}{4},sum_{n=2}^{infty}frac{(-1)^n}{2n+1}=frac{-8+3pi}{12},\
            &&sum_{n=2}^{infty}(-1)^nbigg[frac2{(n+1)^2}-frac2{(n-1)^2}bigg]\
            &=&2sum_{n=1}^{infty}bigg[frac1{(2n+1)^2}-frac1{(2n-1)^2}bigg]-2sum_{n=2}^{infty}bigg[frac1{(2n)^2}-frac1{(2n-2)^2}bigg]\
            &=&-2+frac12=-frac32.
            end{eqnarray*}
            Then you can put them to give the answer.






            share|cite|improve this answer











            $endgroup$


















              6












              $begingroup$

              First note that
              begin{eqnarray*}
              &&frac{1}{n}left[frac{35n-37}{(2n-1)(n-1)^2}+frac{35n+37}{(2n+1)(n+1)^2}right]\
              &=&frac{74}{n}+frac2{(n+1)^2}-frac2{(n-1)^2}+frac{41}{n+1}+frac{41}{n-1}-frac{156}{2n+1}-frac{156}{2n-1}
              end{eqnarray*}
              and
              begin{eqnarray*}
              &&sum_{n=2}^{infty}frac{(-1)^n}{n-1}=ln2,sum_{n=2}^{infty}frac{(-1)^n}{n}=1-ln2,sum_{n=2}^{infty}frac{(-1)^n}{n+1}=-frac12+ln2,\
              &&sum_{n=2}^{infty}frac{(-1)^n}{2n-1}=frac{4-pi}{4},sum_{n=2}^{infty}frac{(-1)^n}{2n+1}=frac{-8+3pi}{12},\
              &&sum_{n=2}^{infty}(-1)^nbigg[frac2{(n+1)^2}-frac2{(n-1)^2}bigg]\
              &=&2sum_{n=1}^{infty}bigg[frac1{(2n+1)^2}-frac1{(2n-1)^2}bigg]-2sum_{n=2}^{infty}bigg[frac1{(2n)^2}-frac1{(2n-2)^2}bigg]\
              &=&-2+frac12=-frac32.
              end{eqnarray*}
              Then you can put them to give the answer.






              share|cite|improve this answer











              $endgroup$
















                6












                6








                6





                $begingroup$

                First note that
                begin{eqnarray*}
                &&frac{1}{n}left[frac{35n-37}{(2n-1)(n-1)^2}+frac{35n+37}{(2n+1)(n+1)^2}right]\
                &=&frac{74}{n}+frac2{(n+1)^2}-frac2{(n-1)^2}+frac{41}{n+1}+frac{41}{n-1}-frac{156}{2n+1}-frac{156}{2n-1}
                end{eqnarray*}
                and
                begin{eqnarray*}
                &&sum_{n=2}^{infty}frac{(-1)^n}{n-1}=ln2,sum_{n=2}^{infty}frac{(-1)^n}{n}=1-ln2,sum_{n=2}^{infty}frac{(-1)^n}{n+1}=-frac12+ln2,\
                &&sum_{n=2}^{infty}frac{(-1)^n}{2n-1}=frac{4-pi}{4},sum_{n=2}^{infty}frac{(-1)^n}{2n+1}=frac{-8+3pi}{12},\
                &&sum_{n=2}^{infty}(-1)^nbigg[frac2{(n+1)^2}-frac2{(n-1)^2}bigg]\
                &=&2sum_{n=1}^{infty}bigg[frac1{(2n+1)^2}-frac1{(2n-1)^2}bigg]-2sum_{n=2}^{infty}bigg[frac1{(2n)^2}-frac1{(2n-2)^2}bigg]\
                &=&-2+frac12=-frac32.
                end{eqnarray*}
                Then you can put them to give the answer.






                share|cite|improve this answer











                $endgroup$



                First note that
                begin{eqnarray*}
                &&frac{1}{n}left[frac{35n-37}{(2n-1)(n-1)^2}+frac{35n+37}{(2n+1)(n+1)^2}right]\
                &=&frac{74}{n}+frac2{(n+1)^2}-frac2{(n-1)^2}+frac{41}{n+1}+frac{41}{n-1}-frac{156}{2n+1}-frac{156}{2n-1}
                end{eqnarray*}
                and
                begin{eqnarray*}
                &&sum_{n=2}^{infty}frac{(-1)^n}{n-1}=ln2,sum_{n=2}^{infty}frac{(-1)^n}{n}=1-ln2,sum_{n=2}^{infty}frac{(-1)^n}{n+1}=-frac12+ln2,\
                &&sum_{n=2}^{infty}frac{(-1)^n}{2n-1}=frac{4-pi}{4},sum_{n=2}^{infty}frac{(-1)^n}{2n+1}=frac{-8+3pi}{12},\
                &&sum_{n=2}^{infty}(-1)^nbigg[frac2{(n+1)^2}-frac2{(n-1)^2}bigg]\
                &=&2sum_{n=1}^{infty}bigg[frac1{(2n+1)^2}-frac1{(2n-1)^2}bigg]-2sum_{n=2}^{infty}bigg[frac1{(2n)^2}-frac1{(2n-2)^2}bigg]\
                &=&-2+frac12=-frac32.
                end{eqnarray*}
                Then you can put them to give the answer.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 31 at 17:36

























                answered Jan 31 at 16:25









                xpaulxpaul

                23.4k24655




                23.4k24655






























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