Evaluating sum using Fourier Series

[tedwillis]

First, I've had to find the Fourier series of F(t) = |sin(t)|, which I've calculated as

f(t) = \frac{2}{\pi} + \sum_{n=1}^{\infty}\frac{4cos(2nt)}{\pi-4\pi n^2}

I'm pretty sure that's right, but now I need to evaluate the sum using the above Fourier series:
\sum_{n=1}^{\infty}\frac{(-1)^n}{4n^2-1}

I don't really have any clue about where to start.

 

[HallsofIvy]

Well, if
f(t)= \frac{2}{\pi}+ \sum_{n= 1}^\infty \frac{4cos(2nt)}{\pi- 4\pi n^2}
it follows that
-\frac{4}{\pi}\sum_{n=1}^\infty \frac{cos(2nt)}{4n^2- 1}= f(t)- \frac{\pi}{2}
so that
\sum_{n=1}^\infty \frac{cos(2nt)}{4n^2- 1}= \frac{\pi}{4}(\frac{\pi}{2}- f(t))

Now, what should you choose t to be?