So I need to use the fourier series of $f(x)=|\sin(x)|$ in $(-\pi,\pi)$ to compute the sums $$\sum_{n\in\mathbb{N}}\frac{1}{4n^2-1}\tag1,$$ $$\sum_{n\in\mathbb{N}}\frac{(-1)^{n+1}}{4n^2-1}.\tag2$$
I correctly obtained the Fourier series to be
$$|\sin(x)|=\frac{2}{\pi}-\frac{4}{\pi}\sum_{n\in\mathbb{N}}\frac{1}{4n^2-1}\cos(2xn)\tag3.$$
To compute sum $(1)$ I used the theorem that states that the series converges to the average of the left and right hand limits of points of discontinuity, like for example $\pi.$ So
$$\frac{|\sin(\pi)|+|\sin(-\pi)|}{2}=0=\frac{2}{\pi}-\frac{4}{\pi}\sum_{n\in\mathbb{N}}\frac{1}{4n^2-1}\cos(2\pi n)\Leftrightarrow (1)=\frac{1}{2},$$
by solving for the sum and since $\cos(2\pi n)=1 \ \forall \ n \ \in \mathbb{N.}$ But I can't apply the same method to compute the sum number $(2).$
does anyone have a suggestion on how one could proceed here?
You could have just used the value of $x=0$ in both sides of the equation for (1). For (2), try inputting $x=\pi/2$ then the left hand side is $1$ and the right hand side is the required summation.