= Solution
The classical <Fourier partial sums> are
$$
s_n^f(t)=\frac{a_0}2+\sum_{j=1}^n[a_j\cos jt+b_j\sin jt],\quad a_j=\frac1\pi\int_{-\pi}^\pi f(t)\cos jt\,dt,\quad b_j=\frac1\pi\int_{-\pi}^\pi f(t)\sin jt\,dt.
$$
Here $a_0=\pi^{-1}\int f$, and functions are extended periodically. One version of the <Dini test for Fourier series> states that if one-sided limits $L_+,L_-$ at $t$ exist and
$$
\int_0^\delta\frac{|f(t+s)+f(t-s)-L_+-L_-|}{s}\,ds<\infty,
$$
then $s_n^f(t)\to(L_++L_-)/2$. A <step function> on a finite interval is constant on each side of any given point for sufficiently small distance, so its displayed numerator is zero there. <Dini test for Fourier series> gives exactly the asserted midpoint of the two one-sided values, including at jumps.
Let $P_n=s_n$ on $L^2$. It is the <orthogonal projection> onto the span of the displayed trigonometric functions through frequency $n$, hence $\|P_n\|\leq1$. For a step-function approximation $\psi$,
$$
\|P_nf-f\|_2\leq\|P_n(f-\psi)\|_2+\|P_n\psi-\psi\|_2+\|\psi-f\|_2\leq2\|f-\psi\|_2+\|P_n\psi-\psi\|_2.
$$
The <convergence on a dense set under a uniform operator bound>, here applied using the allowed convergence for $\psi$ and density of <step functions>, proves \b[$P_nf\to f$ in $L^2$]. Consequently the union of these finite trigonometric spans is dense, which is precisely <completeness> of the <orthonormal system>. It also proves Parseval's equality by passing to the limit in the Pythagorean projection identity.
For $f(t)=t^2$, oddness gives $b_j=0$, direct integration gives $a_0=2\pi^2/3$, and two integrations by parts give $a_j=4(-1)^j/j^2$. Thus
$$
\boxed{t^2=\frac{\pi^2}3+4\sum_{j\geq1}\frac{(-1)^j\cos jt}{j^2}.}
$$
<Dini test for Fourier series> justifies this pointwise on the interval, or absolute coefficient summability gives its continuous extension after the $L^2$ identification. At zero it yields \b[$\sum_{j\geq1}(-1)^{j-1}/j^2=\pi^2/12$]. Parseval gives
$$
\frac{2\pi^5}5=\int_{-\pi}^\pi t^4dt=\frac\pi2\left(\frac{2\pi^2}3\right)^2+16\pi\sum_{j\geq1}\frac1{j^4}.
$$
Therefore \b[$\sum_{j\geq1}j^{-4}=\pi^4/90$].
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