Calculate $sum_{k=0}^{n} frac{(-1)^kk}{4k^2-1}$












2














Calculate $sum_{k=0}^{n} frac{(-1)^k k}{4k^2-1}$



$sum_{k=0}^{n} frac{(-1)^k k}{4k^2-1}=sum_{k=0}^{n} frac{(-1)^k k}{(2k-1)(2k+1)}=sum_{k=0}^{n} (-1)^k kfrac{1}{2}(frac{1}{2k-1}-frac{1}{2k+1})$ after that I get this $sum_{k=0}^{n} (-1)^k frac{k}{2k-1}+(-1)^{n+1}frac{n+1}{2n+1}+sum_{k=0}^{n}frac{1}{2} frac{(-1)^k}{2k+1} $. But that does not help me so much I do not know how to continue, i try everything after I still get the same, can you help me?






discrete-mathematics summation







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  • 1




    Do you need to show your work? Wolframalpha will do this automatically.
    – Ben W
    Nov 25 at 20:18










  • I know but I did not pay so I can not get step by step
    – Marko Škorić
    Nov 25 at 20:19










  • If WolframAlpha just shows you the result, you can then prove it straightforwardly by induction on $n$. And if you want, you can then transform that induction proof into a proof by telescope.
    – darij grinberg
    Nov 25 at 20:22
















2














Calculate $sum_{k=0}^{n} frac{(-1)^k k}{4k^2-1}$



$sum_{k=0}^{n} frac{(-1)^k k}{4k^2-1}=sum_{k=0}^{n} frac{(-1)^k k}{(2k-1)(2k+1)}=sum_{k=0}^{n} (-1)^k kfrac{1}{2}(frac{1}{2k-1}-frac{1}{2k+1})$ after that I get this $sum_{k=0}^{n} (-1)^k frac{k}{2k-1}+(-1)^{n+1}frac{n+1}{2n+1}+sum_{k=0}^{n}frac{1}{2} frac{(-1)^k}{2k+1} $. But that does not help me so much I do not know how to continue, i try everything after I still get the same, can you help me?






discrete-mathematics summation







share|cite|improve this question




















  • 1




    Do you need to show your work? Wolframalpha will do this automatically.
    – Ben W
    Nov 25 at 20:18










  • I know but I did not pay so I can not get step by step
    – Marko Škorić
    Nov 25 at 20:19










  • If WolframAlpha just shows you the result, you can then prove it straightforwardly by induction on $n$. And if you want, you can then transform that induction proof into a proof by telescope.
    – darij grinberg
    Nov 25 at 20:22














2












2








2


1





Calculate $sum_{k=0}^{n} frac{(-1)^k k}{4k^2-1}$



$sum_{k=0}^{n} frac{(-1)^k k}{4k^2-1}=sum_{k=0}^{n} frac{(-1)^k k}{(2k-1)(2k+1)}=sum_{k=0}^{n} (-1)^k kfrac{1}{2}(frac{1}{2k-1}-frac{1}{2k+1})$ after that I get this $sum_{k=0}^{n} (-1)^k frac{k}{2k-1}+(-1)^{n+1}frac{n+1}{2n+1}+sum_{k=0}^{n}frac{1}{2} frac{(-1)^k}{2k+1} $. But that does not help me so much I do not know how to continue, i try everything after I still get the same, can you help me?






discrete-mathematics summation







share|cite|improve this question















Calculate $sum_{k=0}^{n} frac{(-1)^k k}{4k^2-1}$



$sum_{k=0}^{n} frac{(-1)^k k}{4k^2-1}=sum_{k=0}^{n} frac{(-1)^k k}{(2k-1)(2k+1)}=sum_{k=0}^{n} (-1)^k kfrac{1}{2}(frac{1}{2k-1}-frac{1}{2k+1})$ after that I get this $sum_{k=0}^{n} (-1)^k frac{k}{2k-1}+(-1)^{n+1}frac{n+1}{2n+1}+sum_{k=0}^{n}frac{1}{2} frac{(-1)^k}{2k+1} $. But that does not help me so much I do not know how to continue, i try everything after I still get the same, can you help me?






discrete-mathematics summation




discrete-mathematics summation






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share|cite|improve this question













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edited Nov 25 at 20:18

























asked Nov 25 at 20:17









Marko Škorić

70310




70310








  • 1




    Do you need to show your work? Wolframalpha will do this automatically.
    – Ben W
    Nov 25 at 20:18










  • I know but I did not pay so I can not get step by step
    – Marko Škorić
    Nov 25 at 20:19










  • If WolframAlpha just shows you the result, you can then prove it straightforwardly by induction on $n$. And if you want, you can then transform that induction proof into a proof by telescope.
    – darij grinberg
    Nov 25 at 20:22














  • 1




    Do you need to show your work? Wolframalpha will do this automatically.
    – Ben W
    Nov 25 at 20:18










  • I know but I did not pay so I can not get step by step
    – Marko Škorić
    Nov 25 at 20:19










  • If WolframAlpha just shows you the result, you can then prove it straightforwardly by induction on $n$. And if you want, you can then transform that induction proof into a proof by telescope.
    – darij grinberg
    Nov 25 at 20:22








1




1




Do you need to show your work? Wolframalpha will do this automatically.
– Ben W
Nov 25 at 20:18




Do you need to show your work? Wolframalpha will do this automatically.
– Ben W
Nov 25 at 20:18












I know but I did not pay so I can not get step by step
– Marko Škorić
Nov 25 at 20:19




I know but I did not pay so I can not get step by step
– Marko Škorić
Nov 25 at 20:19












If WolframAlpha just shows you the result, you can then prove it straightforwardly by induction on $n$. And if you want, you can then transform that induction proof into a proof by telescope.
– darij grinberg
Nov 25 at 20:22




If WolframAlpha just shows you the result, you can then prove it straightforwardly by induction on $n$. And if you want, you can then transform that induction proof into a proof by telescope.
– darij grinberg
Nov 25 at 20:22










2 Answers
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$$sum_{k=0}^nfrac{(-1)^kk}{4k^2-1}=frac{1}{4}sum_{k=0}^n(-1)^kleft[frac{1}{2k-1}+frac{1}{2k+1}right]$$
$$=frac{1}{4}left[frac{1}{-1}+frac{1}{1}-frac{1}{1}-frac{1}{3}+frac{1}{3}+frac{1}{5}-cdots+frac{(-1)^n}{2n-1}+frac{(-1)^n}{2n+1}right]$$
$$=frac{1}{4}left[frac{1}{-1}+frac{(-1)^n}{2n+1}right]$$






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    1














    HINT



    We have that



    $$sum_{k=0}^nfrac{(-1)^kk}{4k^2-1}=sum_{k=0}^nfrac{(-1)^k}4left(frac{1}{2k+1}+frac{1}{2k-1}right)=$$$$=frac14left(0color{red}{-frac13}-1color{red}{+frac15+frac13-frac17-frac15+frac19+frac17}-ldotsright) $$






    share|cite|improve this answer





















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      2 Answers
      2






      active

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      2 Answers
      2






      active

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      active

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      active

      oldest

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      2














      $$sum_{k=0}^nfrac{(-1)^kk}{4k^2-1}=frac{1}{4}sum_{k=0}^n(-1)^kleft[frac{1}{2k-1}+frac{1}{2k+1}right]$$
      $$=frac{1}{4}left[frac{1}{-1}+frac{1}{1}-frac{1}{1}-frac{1}{3}+frac{1}{3}+frac{1}{5}-cdots+frac{(-1)^n}{2n-1}+frac{(-1)^n}{2n+1}right]$$
      $$=frac{1}{4}left[frac{1}{-1}+frac{(-1)^n}{2n+1}right]$$






      share|cite|improve this answer


























        2














        $$sum_{k=0}^nfrac{(-1)^kk}{4k^2-1}=frac{1}{4}sum_{k=0}^n(-1)^kleft[frac{1}{2k-1}+frac{1}{2k+1}right]$$
        $$=frac{1}{4}left[frac{1}{-1}+frac{1}{1}-frac{1}{1}-frac{1}{3}+frac{1}{3}+frac{1}{5}-cdots+frac{(-1)^n}{2n-1}+frac{(-1)^n}{2n+1}right]$$
        $$=frac{1}{4}left[frac{1}{-1}+frac{(-1)^n}{2n+1}right]$$






        share|cite|improve this answer
























          2












          2








          2






          $$sum_{k=0}^nfrac{(-1)^kk}{4k^2-1}=frac{1}{4}sum_{k=0}^n(-1)^kleft[frac{1}{2k-1}+frac{1}{2k+1}right]$$
          $$=frac{1}{4}left[frac{1}{-1}+frac{1}{1}-frac{1}{1}-frac{1}{3}+frac{1}{3}+frac{1}{5}-cdots+frac{(-1)^n}{2n-1}+frac{(-1)^n}{2n+1}right]$$
          $$=frac{1}{4}left[frac{1}{-1}+frac{(-1)^n}{2n+1}right]$$






          share|cite|improve this answer












          $$sum_{k=0}^nfrac{(-1)^kk}{4k^2-1}=frac{1}{4}sum_{k=0}^n(-1)^kleft[frac{1}{2k-1}+frac{1}{2k+1}right]$$
          $$=frac{1}{4}left[frac{1}{-1}+frac{1}{1}-frac{1}{1}-frac{1}{3}+frac{1}{3}+frac{1}{5}-cdots+frac{(-1)^n}{2n-1}+frac{(-1)^n}{2n+1}right]$$
          $$=frac{1}{4}left[frac{1}{-1}+frac{(-1)^n}{2n+1}right]$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 25 at 20:30









          Ben W

          1,426513




          1,426513























              1














              HINT



              We have that



              $$sum_{k=0}^nfrac{(-1)^kk}{4k^2-1}=sum_{k=0}^nfrac{(-1)^k}4left(frac{1}{2k+1}+frac{1}{2k-1}right)=$$$$=frac14left(0color{red}{-frac13}-1color{red}{+frac15+frac13-frac17-frac15+frac19+frac17}-ldotsright) $$






              share|cite|improve this answer


























                1














                HINT



                We have that



                $$sum_{k=0}^nfrac{(-1)^kk}{4k^2-1}=sum_{k=0}^nfrac{(-1)^k}4left(frac{1}{2k+1}+frac{1}{2k-1}right)=$$$$=frac14left(0color{red}{-frac13}-1color{red}{+frac15+frac13-frac17-frac15+frac19+frac17}-ldotsright) $$






                share|cite|improve this answer
























                  1












                  1








                  1






                  HINT



                  We have that



                  $$sum_{k=0}^nfrac{(-1)^kk}{4k^2-1}=sum_{k=0}^nfrac{(-1)^k}4left(frac{1}{2k+1}+frac{1}{2k-1}right)=$$$$=frac14left(0color{red}{-frac13}-1color{red}{+frac15+frac13-frac17-frac15+frac19+frac17}-ldotsright) $$






                  share|cite|improve this answer












                  HINT



                  We have that



                  $$sum_{k=0}^nfrac{(-1)^kk}{4k^2-1}=sum_{k=0}^nfrac{(-1)^k}4left(frac{1}{2k+1}+frac{1}{2k-1}right)=$$$$=frac14left(0color{red}{-frac13}-1color{red}{+frac15+frac13-frac17-frac15+frac19+frac17}-ldotsright) $$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 25 at 20:29









                  gimusi

                  1




                  1






























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