Let $f(x)=x(\pi-x)$, $x\in (0,\pi)$.
The function satisfies the Dirichlet conditions so its Fourier series, $S_f$ converges pointwise to $f$.
The definition of a Fourier series of $f$ on $[a,a+L]$ is:
$$S(x)=\frac{a_0}{2}+\sum_{n=1}^\infty a_n\cos{ \frac{2n\pi x}{L}} + b_n \sin { \frac{2n\pi x}{L}} $$
with $$a_n = \frac{2}{L}\int_{a}^{a+L} f(x) \cos { \frac{2n\pi x}{L}}dx$$ $$b_n = \frac{2}{L}\int_{a}^{a+L} f(x) \sin { \frac{2n\pi x}{L}}dx$$
With $L=\pi$, $a=0$, this becomes:
$$a_n = \frac{2}{\pi}\int_{0}^{\pi} f(x) \cos(2n x) \ dx = -\frac{1}{n^2}$$ $$b_n = \frac{2}{\pi}\int_{0}^{\pi} f(x) \sin(2n x) \ dx = 0$$
$$a_0 = \frac{\pi^2}{3}$$
Thus
$$S (x) = \frac{\pi^2}{6} - \sum_{n=1}^\infty \frac{\cos(2nx)}{n^2}$$
So, if I denote $$S_n (x) = \frac{\pi^2}{6} - \sum_{k=1}^n \frac{\cos(2kx)}{k^2}$$
I have $S_n \rightarrow f$ pointwise, that is $$\forall x\in[0,\pi], \ f(x) = S(x) = \lim_{n \rightarrow \infty} S_n (x).$$
Now I have to decide whether the convergence is uniform. I would do this by looking at the limiting behaviour of $$\sup_{x\in[0,\pi]} \vert x(\pi - x) + \sum_{k=1}^n \frac{\cos(2kx)}{k^2} - \frac{\pi^2}{6} \vert $$
but I have no idea what to do with this expression. Is there a simpler way to determine uniform convergence?
To prove that an infinite sum of functions $\sum_n f_n(x)$ converges uniformly, all that is needed is to show that the "tails" get small independent of $x$. Here, you'd want to show that, given $\epsilon>0$, there is $N$ such that for $m,n\ge N$, $\sum_{m\le k\le n} |\cos(2kx)/k^2|\;<\;\epsilon$. That is, there's no necessity of referring to the (known) limit to prove uniform convergence.