Find the sum of an alternating, non-geometric series 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.
 A: 
This is another "extreme overkill."

Recalling that the Taylor Series for the arctangent function is given by
$$\arctan(x)=\sum_{n=0}\frac{(-1)^{n}x^{2n+1}}{2n+1}$$
we find that 
$$\begin{align}
\sum_{n=1}^\infty \frac{4n(-1)^n}{4n^2-1}&=\sum_{n=1}^\infty \left(\frac{(-1)^{n}}{2n+1}+\frac{(-1)^{n}}{2n-1}\right)\\\\
&=\left(\arctan(1)-1\right)-\arctan(1)\\\\
&=-1
\end{align}$$
A: 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 $$2\sum_{n\geq1}\frac{\left(-1\right)^{n}}{4n^{2}-1}=\sum_{n\in\mathbb{Z}}\frac{\left(-1\right)^{n}}{4n^{2}-1}+1.$$
A: It is in fact telescopic. For the partial sum:
$$ S_N = \sum_{n=1}^N (-1)^n \left( \frac{1}{2n-1}+\frac{1}{2n+1}\right)=
\sum_{n=1}^N \frac{(-1)^n}{2n-1} - \sum_{k=2}^{N+1} \frac{(-1)^k}{2k-1}
 =-1+\frac{(-1)^N}{2N+1}\rightarrow -1$$
A: It is the moment for the extreme overkill. By differentiation under the integral sign we have that
$$ \sum_{n\geq 1}\frac{4n\cos(\pi n x)}{4n^2-1}\tag{1} $$
over the interval $x\in(0,2)$ is the Fourier cosine series of
$$ f(x) = -1+\cos\left(\frac{\pi x}{2}\right)\cdot\log\cot\left(\frac{\pi x}{4}\right)\tag{2} $$
hence by evaluating $(2)$ at $x=1$ 
$$ \sum_{n\geq 1}\frac{4n(-1)^n}{4n^2-1}=\color{red}{-1}\tag{3} $$
follows.
A: Partial fractions works:
$$\frac{(-1)^n4n}{4n^2-1}=\frac{(-1)^{n}}{2n+1}-\frac{(-1)^{n-1}}{2n-1}$$
This is a telescoping sum of the form:
$$\sum_{n=1}^{x} \left(a_{n+1}-a_{n}\right)=a_{x+1}-a_1$$
With:
$$a_{n}=\frac{(-1)^{n-1}}{2n-1}$$
