How to find the sum of the Fourier series defined by:

$$a\_0 = \frac{\pi^2}{24}$$

$$a\_n = \frac{1}{4\pi n^3}\left((\pi^2n^2-8)\sin(n\pi/2)+4\pi n\cos(n\pi/2)\right)$$
$$b\_n = \frac{1}{4\pi n^3}\left((8-\pi^2n^2)\cos(n\pi/2)+4\pi n\sin(n\pi/2)-8\right),$$
for $x\in [13\pi,15\pi]$. This terms were obtained from
$$a\_n = \frac{1}{\pi}\int\_0^{\pi/2}x^2\cos(nx)dx,~b\_n = \frac{1}{\pi}\int\_0^{\pi/2}x^2\sin(n x)dx. $$

I started by observing that if $n=2k+1,$ then,
$$a\_n = \frac{1}{4\pi n^3}(\pi^2n^2-8)(-1)^{2k+1}$$
$$b\_n = \frac{1}{4\pi n^3}(4\pi n (-1)^{2k+1} -8)$$
for $k = 0,1,2,\ldots.$

Moreover, if $n = 2k$, then,
$$a\_n = \frac{(-1)^{k}}{n^2}$$
$$b\_n = \frac{8-\pi^2n^2}{4\pi n^3}(-1)^{k}$$

So the series is reduced to
$$S = \frac{\pi^2}{48} + \sum\_{k=0}^{\infty}\left(\frac{(-1)^k}{4k^2}\cos(2kx) + \frac{8-\pi^24k^2}{32\pi k^3}\sin(2kx)\right) + \sum\_{k=0}^{\infty}\left(\frac{(\pi^2(2k+1)^2-8)(-1)^{2k+1}}{4\pi (2k+1)^3}\cos((2k+1)x) + \frac{4\pi (2k+1)(-1)^{2k+1}-8}{4\pi(2k+1)^3}\sin((2k+1)\pi x)\right).$$

From here I don't know how to proceed.