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.
Some ideas:
By the given terms, we're talking of a 2$\pi\,-$ periodic function ( since, for example, the Fourier series is given in terms of $\;\cos nx=\cos\frac{n\pi x}L\implies L=\pi,\,T=2\pi\;$, and the same fo the sines terms ), and thus we can translate the interval from $\;[13\pi,\,15\pi]\;$ by $\;2\pi\cdot (-7=)-14\pi\;$ to the much more nice interval $\;[-\pi,\,\pi]\;$
As the Fourier coefficients fulfill
$$\lim_{n\to\infty} a_n=\lim_{n\to\infty} b_n=0\;{}{}(\text{Riemann-Lebesgue Lemma})$$
but not
$$\lim_{n\to\infty} na_n=\lim_{n\to\infty} nb_n=0$$
we have here a function $\;f:[-\pi,\pi]\to\Bbb R\;$ which is $\;2\pi\,-$ periodic, but not continuous on its whole periodic continuation, or with a derivative which is not piecewise continuous. As the given function which expresses $\;a_n, b_n\;$ as integrals is $\;x^2\;$ (and thus undoubtedly with a piecewise continuous derivative over the basic interval $\;[-\pi, \pi)\;$), we can safely deduce the function is not continuous on its periodic continuation, meaning $\;f(-\pi)\neq f(\pi)\;$ ...
Finally, by what we're being told the Fourier coefficients are given from, we have that
$$f(x)=\begin{cases}x^2,&0<x<\frac\pi2\\{}\\ 0,&-\pi<x\le0\,,\,\,\frac\pi2\le x<\pi\end{cases}$$
With the above function, we get exactly the definition of $\;a_n,\,b_n\;$ you were given, and for example
$$a_0=\frac1\pi\int_{-\pi}^\pi f(x)\,dx=\frac1\pi\int_0^{\pi/2} x^2\,dx=\left.\frac1{3\pi}x^3\right|_0^{\pi/2}=\frac1{3\pi}\frac{\pi^3}8=\frac{\pi^2}{24}\implies \frac{a_0}2=\frac{\pi^2}{48}$$
and we indeed get the first Fourier coefficient as given...
By Dirichlet Theorem, the Fourier series converge pointwise to the function at every continuity point of it, and to the average value of the function where it is discontinue. In symbols:
$$\frac{f(x^+)+f(x^-)}2= \text{ Fourier Series}$$
Because we don't have a continue function over its periodic continuation, the above convergence cannot be uniform.