Disclaimer: I am only to work within the orthonormal system $\{e^{inx}\mid n\in\mathbb{Z}\}$! No cosine series/sine series straight off the bat.
I am working with the function $f(x)=\begin{cases} 0 & -\pi< x<0 \\\sin x & 0\leq x\leq \pi \end{cases}$
I am to show that $\sum_{-\infty}^{\infty}\frac{1}{4k^2-1}=0$ and $\sum_{-\infty}^{\infty}\frac{1}{(4k^2-1)^2}=\frac{\pi^2}{8}$
I have started off by finding the Fourier coefficients, which are:
$$c_1=-\frac{i}{4}, \hspace{0.5cm} c_{-1}=\frac{i}{4}, \hspace{0.5cm} c_0=\frac{1}{\pi},\hspace{0.5cm} c_{2k-1}=0, k\in\mathbb{Z}\setminus\{0,1\} , \hspace{0.5cm}c_{2k}=-\frac{1}{\pi}\frac{1}{4k^2-1}$$
I can see, that the above sums are quite close to summing over $(-\pi )c_{2k}$ and $(-\pi)^2c_{2k}^2$ respectively. I have found the Fourier series of $f$ to be
$$c_{-1}+c_0+c_1+\sum_{n=2}^{\infty}c_n(e^{inx}+e^{-inx})=c_0+2\sum_{n=2}^{\infty}c_n\cos(nx)$$
I have previously shown, that the Fourier series converges to f, so since $f(0)=0$, I evaluate the Fourier series in $0$ and get:
$$0=f(0)=\frac{1}{\pi}-\sum_{n=2}^\infty\frac{1}{\pi}\frac{1}{2k^2-1}$$
I feel like I might have miscalculated something, somewhere along the line or overseen something. However, I note that every uneven term of the series is $0$, which I am unsure of how to use.
As for the second sum, I am thinking that I should apply Parseval's Identity, seeing as it has something that is nearly my Fourier coefficients squared?
Bottom line is, I would like some hints or guidance to show the equalities presented at the top, using the Fourier series of $f$.
Thank you.