$f(x)=|x+\frac{\pi}{2}|-2|x|+|x-\frac{\pi}{2}|$
After using Fourier transformation I get
$a_0 = \frac{\pi}{2}$
$a_n = \frac{8}{\pi n^2}sin^2(\frac{n\pi}{4})$
$f(x)= \frac{a_0}{2} + \sum_{n=1} ^{\infty} a_n*cos(nx)$
After this of I substitute n=2k and k=2h+1 I can get values but not for the sum in question. Is it even possible to solve this?
I suppose your calculation is true. You found $$f(x)= |x+\frac{\pi}{2}|-2|x|+|x-\frac{\pi}{2}| =\frac{\pi}{4} + \sum_{n=1} ^{\infty} \frac{8}{\pi n^2}\sin^2(\frac{n\pi}{4})\cos(nx)$$ let $x=0$ then $$\frac{3\pi}{4} = \sum_{n=1} ^{\infty} \frac{8}{\pi n^2}\dfrac12\left(1-\cos\frac{n\pi}{2}\right)$$ or $$\frac{3\pi^2}{16} = \sum_{n=1} ^{\infty} \frac{1}{n^2}\left(1-\cos\frac{n\pi}{2}\right)$$ we may let $n=4k, 4k+1, 4k+2, 4k+3$ and after simplification, only terms with $n=4k+2$ remain. Can you proceed from here?
