How to use a fourier series to find the sums..

Question

So i have a question which asks to find the fourier series of $\left\vert\,\sin\left(x\right)\,\right\vert\,$. I have worked out the solution as $$ {2 \over\pi} - {4 \over \pi}\sum_{k = 1}^{\infty}{\cos\left(2kx\right) \over 4k^{2} - 1} $$

Which i am pretty sure is correct as i have the solution in my book.

The second part of the question asks to work out the sums of $$ \sum_{k = 1}^{\infty}{1 \over 4k^{2} - 1}\qquad\mbox{and}\qquad \sum_{k = 1}^{\infty}{\left(-1\right)^{k} \over 4k^{2} - 1} $$

Im sure this is probably very simple but i have no solution for this and I am struggling to search for an explanation of how to do this on google. Could someone please tell me know it is done ?.

Many thanks.

Answer 1

Well, if the fourier series converges to the function, plug in zero as the argument to $|\sin x|$ and to the fourier series. See what happens. In the second instance, plug in $\pi/2$ as argument.

Answer 2

Solve for the infinite sum:

$$ \sum_{k=1}^\infty \frac{\cos(2nx)}{4n^2-1} = \frac{1}{2} - \frac{\pi}{4} \left| \sin x \right| $$

What happens if you let $x = 1$?

What happens if you let $x = \frac{\pi}{2}$?

Answer 3

what a beautiful result! and if one plugs in $\frac{\pi}4$ it appears one gets (something like!) $$\frac{\sqrt{2}}2=\frac2{\pi} \left(1+\sum_{n=1} \frac{(-1)^{n+1}}{(4n+1)(4n-1)} \right)$$