Finding Fourier series and evaluating at a point

Question

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:

Answer 1

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!