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compute the sum $\sum_1^{+\infty}(-1)^n\frac{n}{4n^2-1}$

I can only see that the summand is odd, hence $\sum_{-\infty}^{+\infty}(-1)^n\frac{n}{4n^2-1}=0$

I know there is a standard way of computing summation of series using residue formula, but i can't see how to apply that in this case

1 Answers 1

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$\frac r{4r^2-1}=\frac r{(2r-1)(2r+1)}=\frac14\frac{(2r+1)+(2r-1)}{(2r-1)(2r+1)}=\frac14\left(\frac1{2r-1}+\frac1{2r+1}\right)$

$\implies 4\sum_1^n(-1)^r\frac r{4r^2-1}=\sum_1^n(-1)^r\left(\frac1{2r-1}+\frac1{2r+1}\right)$

$=-1-\frac13+\frac13+\frac15-\frac15-\frac17+\cdots +(-1)^{n-1}\frac1{2n-3}+(-1)^{n-1}\frac1{2n-1}+(-1)^n\frac1{2n-1}+(-1)^n\frac1{2n+1}$ $=-1+(-1)^n\frac1{2n+1}$

So, setting $n\to \infty, \sum_1^{\infty}(-1)^r\frac r{4r^2-1}=-\frac14$