Evaluating $\sum_{n=1}^{\infty} \frac{4(-1)^n}{1-4n^2}$ I recently found a series representation for 1 from the calculation of a Fourier series:
$$1 = \frac{2}{\pi} + \sum_{n=1}^{\infty} \frac{4(-1)^n}{\pi(1-4n^2)}$$
From this, I can easily find that
$$\sum_{n=1}^{\infty} \frac{4(-1)^n}{1-4n^2} = \pi - 2,$$
but what other methods are there to evaluate the sum? I don't even know where to start.

$$\frac{4}{1-4 n^2} = \frac{2}{1-2 n} + \frac{2}{1+2 n}$$
Thus,
$$\begin{align}\sum_{n=1}^{\infty} \frac{4 (-1)^n}{1-4 n^2} &= 2 \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{2 n-1}+2 \sum_{n=1}^{\infty} \frac{(-1)^n}{2 n+1} \\ &= 2 \frac{\pi}{4} + 2 \left (\frac{\pi}{4}-1 \right ) \\ &= \pi-2 \end{align}$$
