TITLE: Find the sum of an alternating, non-geometric series
QUESTION [11 upvotes]: Looking to the following series:
$$\sum_{n=1}^\infty \frac{(-1)^n(4n)}{4n^2-1}$$
It converges according to Leibniz criteria. However it does not seem to be a telescopic series (if you take partial fractions, you end up with two positive terms), neither a geometric one, so I can not figure out a way to find its sum. Maybe I am missing something here. 
Thanks for your time and I appreciate any help.

REPLY [7 votes]: Third overkill. We have $$\sum_{n\geq1}\frac{\left(-1\right)^{n}4n}{4n^{2}-1}=2\sum_{n\geq1}\frac{\left(-1\right)^{n}2n-1+1}{4n^{2}-1}
 $$ $$=2\sum_{n\geq1}\frac{\left(-1\right)^{n}}{2n+1}+2\sum_{n\geq1}\frac{\left(-1\right)^{n}}{4n^{2}-1}
 $$ $$=\frac{\pi}{2}-2+1-\frac{\pi}{2}=\color{red}{-1}
 $$ where the first sum follows from the Taylor series of the $\arctan
 $ function and the second follows from the summation formula $$\sum_{n\in\mathbb{Z}}\left(-1\right)^{n}f\left(n\right)=-\sum\left\{ \textrm{residues of }\pi\csc\left(\pi z\right)f(z)\textrm{ at }f\left(z\right)\textrm{ poles}\right\} 
 $$ observing that $$2\sum_{n\geq1}\frac{\left(-1\right)^{n}}{4n^{2}-1}=\sum_{n\in\mathbb{Z}}\frac{\left(-1\right)^{n}}{4n^{2}-1}+1.$$