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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.

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    The series can be evaluated more easily using a telescoping series, in case you are interested.2012-10-25

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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.

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    There are examples of uniformly continuous functions whose Fourier series diverge on a dense set. If you have Katznelson's book he describes this in the remark after the proof of Theorem 2.$1$. More generally, pointwise convergence behavior can be relevant due to such things as Gibbs' Phenomenon.2013-04-07