Starting with Fourier cosine series for $f(x)=x^2$ in the interval $(0,l),$ use Parseval identity to compute $\sum_{n=1}^{\infty}\frac{1}{n^4}$.
Attempt: I have that the coefficient $A_n$ for cosine series is $$A_n=(1/(n^3 \pi^3))2 l^2 (2 n \pi \cos( n \pi) + (-2 + n^2 \pi^2) \sin(n \pi))$$
By other hand I have that $\int_0^lx^2dx<\infty$ so the convergence is in $L^2$ sense, so I have the parseval identity, i.e.
$\frac{l^5}{5}=\int_0^l|x^2|^2dx=\sum_{n=1}^\infty A_n^2\Vert \cos\frac{n\pi x}{l}\Vert^2=\sum_{n=1}^\infty A_n^2 (l (n \sin(2 \pi n) + n \sinh(2 \pi n)))/(4 \pi n^2)$
Now I have to take out the $\frac{1}{n^4}$ term from the series, which is definetely not an easy task.
Am I correct so far?
Is there a easy way to proceed? If not, Is there an alternative (easier) solution to this exercise?
Edit
Noticing the cosine multiples of $\pi$ I get
$\frac{\pi^5}{20}=\sum_{n=1}^{\infty}\frac{\sinh(2\pi n)}{n^5}$
but this doesn't looks like $\sum_{n=1}^{\infty}\frac{1}{n^4}$ at all... Unless $\frac{\sinh(2\pi n)}{n}=1$
What am I doing wrong?
Please help me.
