Finding Fourier series and evaluating at a point Ive been asked to find the fourier series of the following 
Need help with finishing the question, i have included my work so far below:

and following that i need help with the final part of the question:

You wrote the correct expression for the coefficients $a_n$, $n\ge1$.  However, there must have been an error in your arithmetic for carrying out the integral
$$a_n=\frac1\pi\int_{-\pi}^\pi (x^2-\pi^2)^2\cos(nx)\,dx=\frac{48(-1)^{n-1}}{n^4}$$
Also, we need the term $a_0$, which is
$$a_0=\frac1{2\pi}\int_{-\pi}^\pi (x^2-\pi^2)^2\,dx=\frac{8\pi^4}{15}$$
So, we have
$$\begin{align}
f(x)&=(x^2-\pi^2)^2\\\\
&=\frac{8\pi^4}{15}-\sum_{n=1}^\infty \frac{48(-1)^{n}}{n^4}\cos(nx)
\end{align}$$
Now, for $x=\pi$, $f(x)=0$. Furthermore, $\cos(n\pi)=(-1)^n$.  Hence, we have
$$\begin{align}
f(\pi)&=0\\\\
&=\frac{8\pi^4}{15}-\sum_{n=1}^\infty \frac{48(-1)^{n}}{n^4}\cos(n\pi)\\\\
&=\frac{8\pi^4}{15}-\sum_{n=1}^\infty \frac{48(-1)^{n}}{n^4}(-1)^n\\\\
0&=\frac{8\pi^4}{15}-\sum_{n=1}^\infty \frac{48}{n^4}\tag1
\end{align}$$
From $(1)$, we see that
$$\sum_{n=1}^\infty \frac{1}{n^4}=\frac{\pi^4}{90}$$
as was to be determined!
