I was solving this question I saw in a textbook. The question is :
Calculate the Fourier series for $ f(x) = |\sin x| $ for $-\pi \leq x \leq \pi$.
Which I had $ f(x) = \frac{a_{0}}{2} + \sum a_{n} \cos nx$, where $a_{n} = \frac{2}{\pi} \int_{0}^{\pi} |\sin x|\cos nx dx$. Which I have been able to do; that is by using trig substitution. I had $$ \frac{(n-1)[(-1)^{n+1} - 1] + [(-1)^{n+1} -1)](n+1)}{\pi (n^2 - 1)}$$
For the convergence of $f(x)$, I know it convergences at $x = 0$ because the function is even continuous function.That is by using $$\frac{f(-\pi) + f(\pi)}{2}.$$ Now the problem is, how do I use the Fourier series in above to show that $ \sum_{1}^{\infty} \frac{1}{4n^{2} - 1} = \frac{1}{2}$. I really need guidelines.
Use the fact that $\sin(x)$ is nonnegative on $0 \leq x \leq \pi$, so that your $a_n$ is given by $$a_n = {2 \over \pi}\int_0^{\pi}\sin(x)\cos(nx)\,dx$$ For $n=0$, compute this directly. Otherwise use $\sin(a)\cos(b) = {\sin(a + b) + \sin(a-b) \over 2}$. The Fourier series converges to $|\sin(x)|$ everywhere because it's piecewise continuously differentiable. Plug in $x = 0$ into the Fourier series to get the summation.