How to find the value $sum_{ngeq2}^{infty}(-1)^{n+1}frac{n}{n^2-1}$ converges to?












1












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How to find the value this sum converges to?$$sum_{ngeq2}^{infty}(-1)^{n+1}frac{n}{n^2-1} $$
I've tried writing it like this
$$sum_{ngeq2}^{infty}(-1)^{n+1}·n·Bigg(frac{1/2}{n-1}-frac{1/2}{n+1}Bigg) $$
and writing a few terms, but they won't cancel and I ended up with no ideas, any hint?
I haven't learnt integration nor differentiation FYI.






real-analysis sequences-and-series summation







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  • $begingroup$
    Ok I miss wrote one thing, thank guys.
    $endgroup$
    – iggykimi
    Jan 24 at 16:58
















1












$begingroup$


How to find the value this sum converges to?$$sum_{ngeq2}^{infty}(-1)^{n+1}frac{n}{n^2-1} $$
I've tried writing it like this
$$sum_{ngeq2}^{infty}(-1)^{n+1}·n·Bigg(frac{1/2}{n-1}-frac{1/2}{n+1}Bigg) $$
and writing a few terms, but they won't cancel and I ended up with no ideas, any hint?
I haven't learnt integration nor differentiation FYI.






real-analysis sequences-and-series summation







share|cite|improve this question











$endgroup$












  • $begingroup$
    Ok I miss wrote one thing, thank guys.
    $endgroup$
    – iggykimi
    Jan 24 at 16:58














1












1








1





$begingroup$


How to find the value this sum converges to?$$sum_{ngeq2}^{infty}(-1)^{n+1}frac{n}{n^2-1} $$
I've tried writing it like this
$$sum_{ngeq2}^{infty}(-1)^{n+1}·n·Bigg(frac{1/2}{n-1}-frac{1/2}{n+1}Bigg) $$
and writing a few terms, but they won't cancel and I ended up with no ideas, any hint?
I haven't learnt integration nor differentiation FYI.






real-analysis sequences-and-series summation







share|cite|improve this question











$endgroup$




How to find the value this sum converges to?$$sum_{ngeq2}^{infty}(-1)^{n+1}frac{n}{n^2-1} $$
I've tried writing it like this
$$sum_{ngeq2}^{infty}(-1)^{n+1}·n·Bigg(frac{1/2}{n-1}-frac{1/2}{n+1}Bigg) $$
and writing a few terms, but they won't cancel and I ended up with no ideas, any hint?
I haven't learnt integration nor differentiation FYI.






real-analysis sequences-and-series summation




real-analysis sequences-and-series summation






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edited Jan 24 at 17:32









Lorenzo B.

1,8602520




1,8602520










asked Jan 24 at 16:39









iggykimiiggykimi

30710




30710












  • $begingroup$
    Ok I miss wrote one thing, thank guys.
    $endgroup$
    – iggykimi
    Jan 24 at 16:58


















  • $begingroup$
    Ok I miss wrote one thing, thank guys.
    $endgroup$
    – iggykimi
    Jan 24 at 16:58
















$begingroup$
Ok I miss wrote one thing, thank guys.
$endgroup$
– iggykimi
Jan 24 at 16:58




$begingroup$
Ok I miss wrote one thing, thank guys.
$endgroup$
– iggykimi
Jan 24 at 16:58










2 Answers
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Note that$$(-1)^{n+1}frac n{n^2-1}=frac12timesfrac{(-1)^{n+1}}{n-1}+frac12timesfrac{(-1)^{n+1}}{n+1}.$$But$$sum_{n=2}^inftyfrac{(-1)^{n+1}}{n-1}=sum_{n=1}^inftyfrac{(-1)^n}n=-log(2)$$and$$sum_{n=2}^inftyfrac{(-1)^{n+1}}{n+1}=sum_{n=3}^inftyfrac{(-1)^n}n=left(sum_{n=1}^inftyfrac{(-1)^n}nright)+1-frac12=-log(2)+frac12.$$






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    Try writing $$begin{align}nleft({1over n-1}-{1over n+1}right)&=
    nleft({1over n-1}-{1over n} + {1over n} -{1over n+1}right)\
    &=nleft({1over n(n-1)}+{1over n(n+1)}right)\&={1over n-1}+{1over n+1}end{align}$$






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      2 Answers
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      2 Answers
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      4












      $begingroup$

      Note that$$(-1)^{n+1}frac n{n^2-1}=frac12timesfrac{(-1)^{n+1}}{n-1}+frac12timesfrac{(-1)^{n+1}}{n+1}.$$But$$sum_{n=2}^inftyfrac{(-1)^{n+1}}{n-1}=sum_{n=1}^inftyfrac{(-1)^n}n=-log(2)$$and$$sum_{n=2}^inftyfrac{(-1)^{n+1}}{n+1}=sum_{n=3}^inftyfrac{(-1)^n}n=left(sum_{n=1}^inftyfrac{(-1)^n}nright)+1-frac12=-log(2)+frac12.$$






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        $begingroup$

        Note that$$(-1)^{n+1}frac n{n^2-1}=frac12timesfrac{(-1)^{n+1}}{n-1}+frac12timesfrac{(-1)^{n+1}}{n+1}.$$But$$sum_{n=2}^inftyfrac{(-1)^{n+1}}{n-1}=sum_{n=1}^inftyfrac{(-1)^n}n=-log(2)$$and$$sum_{n=2}^inftyfrac{(-1)^{n+1}}{n+1}=sum_{n=3}^inftyfrac{(-1)^n}n=left(sum_{n=1}^inftyfrac{(-1)^n}nright)+1-frac12=-log(2)+frac12.$$






        share|cite|improve this answer









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          4












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          4





          $begingroup$

          Note that$$(-1)^{n+1}frac n{n^2-1}=frac12timesfrac{(-1)^{n+1}}{n-1}+frac12timesfrac{(-1)^{n+1}}{n+1}.$$But$$sum_{n=2}^inftyfrac{(-1)^{n+1}}{n-1}=sum_{n=1}^inftyfrac{(-1)^n}n=-log(2)$$and$$sum_{n=2}^inftyfrac{(-1)^{n+1}}{n+1}=sum_{n=3}^inftyfrac{(-1)^n}n=left(sum_{n=1}^inftyfrac{(-1)^n}nright)+1-frac12=-log(2)+frac12.$$






          share|cite|improve this answer









          $endgroup$



          Note that$$(-1)^{n+1}frac n{n^2-1}=frac12timesfrac{(-1)^{n+1}}{n-1}+frac12timesfrac{(-1)^{n+1}}{n+1}.$$But$$sum_{n=2}^inftyfrac{(-1)^{n+1}}{n-1}=sum_{n=1}^inftyfrac{(-1)^n}n=-log(2)$$and$$sum_{n=2}^inftyfrac{(-1)^{n+1}}{n+1}=sum_{n=3}^inftyfrac{(-1)^n}n=left(sum_{n=1}^inftyfrac{(-1)^n}nright)+1-frac12=-log(2)+frac12.$$







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          answered Jan 24 at 16:51









          José Carlos SantosJosé Carlos Santos

          168k22132236




          168k22132236























              3












              $begingroup$

              Try writing $$begin{align}nleft({1over n-1}-{1over n+1}right)&=
              nleft({1over n-1}-{1over n} + {1over n} -{1over n+1}right)\
              &=nleft({1over n(n-1)}+{1over n(n+1)}right)\&={1over n-1}+{1over n+1}end{align}$$






              share|cite|improve this answer









              $endgroup$


















                3












                $begingroup$

                Try writing $$begin{align}nleft({1over n-1}-{1over n+1}right)&=
                nleft({1over n-1}-{1over n} + {1over n} -{1over n+1}right)\
                &=nleft({1over n(n-1)}+{1over n(n+1)}right)\&={1over n-1}+{1over n+1}end{align}$$






                share|cite|improve this answer









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                  3












                  3








                  3





                  $begingroup$

                  Try writing $$begin{align}nleft({1over n-1}-{1over n+1}right)&=
                  nleft({1over n-1}-{1over n} + {1over n} -{1over n+1}right)\
                  &=nleft({1over n(n-1)}+{1over n(n+1)}right)\&={1over n-1}+{1over n+1}end{align}$$






                  share|cite|improve this answer









                  $endgroup$



                  Try writing $$begin{align}nleft({1over n-1}-{1over n+1}right)&=
                  nleft({1over n-1}-{1over n} + {1over n} -{1over n+1}right)\
                  &=nleft({1over n(n-1)}+{1over n(n+1)}right)\&={1over n-1}+{1over n+1}end{align}$$







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                  share|cite|improve this answer










                  answered Jan 24 at 16:55









                  saulspatzsaulspatz

                  17k31435




                  17k31435






























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