Find the sum of an alternating, non-geometric series

It is in fact telescopic. For the partial sum: $$ S_N = \sum_{n=1}^N (-1)^n \left( \frac{1}{2n-1}+\frac{1}{2n+1}\right)= \sum_{n=1}^N \frac{(-1)^n}{2n-1} - \sum_{k=2}^{N+1} \frac{(-1)^k}{2k-1} =-1+\frac{(-1)^N}{2N+1}\rightarrow -1$$

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.

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:

$$a_{n}=\frac{(-1)^{n-1}}{2n-1}$$