= Solution
With the classical real coefficients
$$
a_j=\frac1\pi\int_{-\pi}^\pi f(x)\cos(jx)dx,\qquad b_j=\frac1\pi\int_{-\pi}^\pi f(x)\sin(jx)dx,
$$
the partial sum is $s_n^f=a_0/2+\sum_{j=1}^n(a_j\cos jx+b_j\sin jx)$. It is the <orthogonal projection> onto the first trigonometric modes. <Completeness> of the trigonometric <orthonormal> system therefore gives $\|s_n^f-f\|_2\to0$.
For $f\in C^1$ periodic, <integration by parts> has zero endpoint terms and gives $a_j(f')=jb_j(f)$, $b_j(f')=-ja_j(f)$, and $a_0(f')=0$. Thus
$$
\boxed{s_n^{f'}(x)=\sum_{j=1}^n[jb_j(f)\cos jx-ja_j(f)\sin jx]=(s_n^f)'(x).}
$$
For periodic $C^2$ functions, another <integration by parts> gives $a_j(f)=-a_j(f'')/j^2$ and $b_j(f)=-b_j(f'')/j^2$. Riemann-Lebesgue makes the numerator sequences tend to zero, in particular bounded. The <Weierstrass M-test> and $\sum j^{-2}<\infty$ prove absolute and <uniform convergence>. The resulting <continuous> sum agrees with $f$ in $L^2$ by the first result. Two <continuous> functions equal almost everywhere are equal everywhere, so the pointwise sum is $f$ for every real $x$.
For the <polynomial> example the periodic extension is $C^2$: the function, its first <derivative> and its second <derivative> match at the endpoints. It is even, so all sine coefficients vanish. Direct integration gives
$$
\frac{a_0}{2}=\frac{7\pi^4}{15},\qquad a_j=\frac{48(-1)^j}{j^4}\quad(j\geq1).
$$
For example, <integration by parts> gives $\int_0^\pi x^2\cos(jx)dx=2\pi(-1)^j/j^2$ and $\int_0^\pi x^4\cos(jx)dx=4\pi^3(-1)^j/j^2-24\pi(-1)^j/j^4$, from which the cancellation of the $j^{-2}$ terms follows. Therefore
$$
\boxed{f(x)=\frac{7\pi^4}{15}+48\sum_{j=1}^\infty\frac{(-1)^j\cos(jx)}{j^4}.}
$$
At $x=0$, the left side vanishes, yielding
$$
\boxed{\sum_{j=1}^\infty\frac{(-1)^{j+1}}{j^4}=\frac{7\pi^4}{720}.}
$$
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