Evaluate $\sum^\infty_{n=1} \frac{1}{n^4} $using Parseval's theorem (Fourier series) 
Evaluate $\sum^\infty_{n=1} \frac{1}{n^4} $using Parseval's theorem (Fourier series).

I have , somehow, to find the sum of $\sum_{n=1}^\infty \frac{1}{n^4}$ using Parseval's theorem.
I tried some things that didn't work so I won't post them.
Can you please explain me how do I find the sum of this series using Parseval's identity?
Thanks

We have $f \in  L^2 \left[ -\pi, \pi \right]$ then 
\begin{align}
 \dfrac{1}{2}{A_0^2} + \displaystyle \sum_{n=1}^{\infty}{\left( A_n^2 + B_n^2 \right)} = \dfrac{1}{\pi} \int_{-\pi}^ \pi {f^2(x)} dx.
 \end{align}
Here $A_n$ and $B_n$ are Fourier coefficients of $f$.
Let $f(x) = x^2$ for $x \in \left( -\pi, \pi \right)$. Then $f \in  L^2 \left[ -\pi, \pi \right]$ and
 $$f(x) \sim \dfrac{{\pi}^2}{3} + 4{\displaystyle \sum_{n =1}^{\infty}{ \dfrac{(-1)^{n}}{n^2}\cos nx}}.$$
Therefore
$$\dfrac{1}{2}\left(\dfrac{2 \pi^2}{3}\right)^2 +\displaystyle\sum_{n=1}^{\infty} \left( \dfrac{4.(-1)^n}{n^2} \right)^2 = \dfrac{1}{\pi} \int_{-\pi}^\pi{x^4}dx $$
So $$  \displaystyle \sum_{n=1}^{\infty} \dfrac{1}{n^4}= \dfrac{\pi^4}{90}.$$
