Fourier series of $f(x) = |x|^3$ and evaluating series I found the Fourier serie for the function $$f: [-\pi, \pi], \quad f(x) = |x|^3$$
Coefficents:
$$
a_0 = \frac{\pi^3}{2}
$$
$$
a_n = \frac{6 \pi}{n^2} \cos(n\pi) - \frac{12}{n^4\pi} \cos(n\pi) + \frac{6}{n^4}
$$
$$
b_n = 0
$$
So, the Fourier serie is given by
$$
f(x) = \frac{a_0}{2} + \sum_{n = 1}^{+\infty} a_n \cos(nx)
$$
$$
f(x) = \frac{\pi^3}{4} + \sum_{n = 1}^{+\infty}\left( \frac{6 \pi}{n^2} (-1)^n - \frac{12}{n^4\pi} (-1)^n + \frac{6}{n^4} \right) \cos(nx)
$$
Now, I should evaluate the following serie:
$$
\sum_{n = 1}^{+\infty} \frac{1}{n^4}(\pi^2 n^2 (-1)^n - 2 (-1)^n + 2)
$$
How can I find it with the help of this Fourier series?

I think the fourier coefficients are
$$n\neq1,\;\;a_n=\frac{6(2-n^2\pi^2)}{\pi n^4}(-1)^n-\frac{12}{\pi n^4}$$
So when we susntitute $\;x=0\;$ , we get (Dirichlet's convergence theorem)
$$0=\frac{\pi^3}4+6\sum_{n=1}^\infty\left(\frac{(2-n^2\pi^2)}{\pi n^4}(-1)^n-\frac{2}{\pi n^4}\right)\implies-\frac{\pi^3}{24}=\sum_{n=1}^\infty\left(\frac{(2-n^2\pi^2)}{\pi n^4}(-1)^n-\frac{2}{\pi n^4}\right)\implies$$
$$-\frac{\pi^4}{24}=\sum_{n=1}^\infty\left(\frac{(2-n^2\pi^2)}{ n^4}(-1)^n-\frac{2}{ n^4}\right)$$
and there you have your sum...(of course, I assume you know the sum of $\;\sum\limits_{n=1}^\infty\frac1{n^4}\;$ ...)
