Fourier Series for f(x) = sin(3x/2) and Evaluating Series for (1/(4n^2-9))^2

[theorem]

Homework Statement

Evaluate following series:
$$\sum_{n=1}^\infty \frac{1}{(4n^2-9)^2}$$
by finding the Fourier series for the $2\pi$-periodic function
$$f(x) = \begin{cases} sin(3x/2) & 0<x<\pi \\ 0 & otherwise \end{cases}$$

Homework Equations

$$a_n = \frac{1}{\pi}\int_{-\pi}^{\pi} f(x)cos(nx)dx = -\frac{6}{\pi(4n^2-9)}$$
$$b_n = \frac{1}{\pi}\int_{-\pi}^{\pi} f(x)sin(nx)dx = \frac{4ncos(\pi n)}{\pi(4n^2-9)}$$
$$f(x) = \frac{1}{2}a_0 + \sum_{n=1}^\infty \left(a_ncos(nx)+b_nsin(nx)\right) \\ = \frac{1}{2}\frac{2}{3\pi} + \frac{1}{\pi}\sum_{n=1}^\infty \left( \frac{4ncos(\pi n)sin(nx)-6cos(nx)}{4n^2-9}\right)$$

The Attempt at a Solution

I have found the Fourier series and evaluated 1/(4n^2-9) as it was the first part of this exercise. However, I am not sure how to evaluate it for the square. I figured I just needed to square the answer but it turns out it's that simple .

I found Parseval's formula in my book
$$\frac{1}{\pi}\int_T |f(x)|^2dx = \frac{1}{2}|a_0|^2 + \sum_{n=1}^\infty (|a_n|^2+|b_n|^2)$$.
which I tried using, but the $b_n$ term gives me an ugly expression in the numerator.

 

[pasmith]

$b_n$ should not be a function of $x$. Take another look at it, and remember that $\cos(\pm n\pi) = (-1)^n$.

 

[theorem]

Oh yes, my mistake. I meant cos(pi n) not cos(xn).