= Solution
The function is even, so its <Fourier series> has $b_n=0$. In the convention $f\sim a_0/2+\sum_{n\geq1}(a_n\cos nx+b_n\sin nx)$,
$$
a_0=\frac2\pi\int_0^\pi x^2\,dx=\frac{2\pi^2}{3}.
$$
Twice applying <integration by parts> gives, for $n\geq1$,
$$
\begin{aligned}
\int_0^\pi x^2\cos(nx)\,dx
&=-\frac2n\int_0^\pi x\sin(nx)\,dx\\
&=\frac{2\pi(-1)^n}{n^2},\qquad a_n=\frac{4(-1)^n}{n^2}.
\end{aligned}
$$
Therefore \b[the periodic Fourier representation is]
$$
\boxed{x^2=\frac{\pi^2}{3}+4\sum_{n=1}^\infty\frac{(-1)^n\cos(nx)}{n^2},\qquad -\pi\leq x\leq\pi,}
$$
with periodic continuation. The periodic extension is continuous, including the identified endpoints, and piecewise smooth, so the <Fourier series> converges to it. The coefficients also give <absolute convergence> and <uniform convergence> by comparison with $\sum n^{-2}$.
At $x=\pi$, the cosine cancels the alternating coefficient sign; at $x=0$, it is identically one. These evaluations yield
$$
\boxed{\sum_{n=1}^\infty\frac1{n^2}=\frac{\pi^2}{6},\qquad \sum_{n=1}^\infty\frac{(-1)^{n+1}}{n^2}=\frac{\pi^2}{12}.}
$$
The first identity solves the <Basel problem>.
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