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The function is even, so its Fourier series has $b_n=0$. In the convention $f\sim a_0/2+{\sum }_{n\ge 1}(a_n\cos nx+b_n\sin nx)$,Twice applying integration by parts gives, for $n\ge 1$,Therefore the periodic Fourier representation iswith 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 yieldThe first identity solves the Basel problem.

The displayed definition uses iterated Eisenstein summation in weight two: the sum in $n$ is evaluated before the sum in $m$, with the single $(0,0)$ term omitted. This order is essential. The two-dimensional lattice series does not have absolute convergence, so arbitrary rearrangement would not be justified.

For noninteger $w$, the cosecant partial-fraction identity isFor completeness, apply the residue theorem to $\pi \cot (\pi \zeta )/(\zeta -w{)}^2$ on squares with large half-integer sides. The cotangent is bounded on the contours and the integral is $O(R^{-1})$. Its residues at the integers are $(n-w{)}^{-2}$ and its residue at $w$ is $-{\pi }^2{\csc }^2(\pi w)$, proving the formula. If $\mathrm{Im}w>0$, the geometric-series expression $\cot (\pi w)=-i(1+2{\sum }_{r\ge 1}{e}^{2\pi irw})$, differentiated termwise, gives the cotangent partial-fraction Fourier kernel

Put $q=e^{2\pi iz}$ with $z$ in the complex upper half-plane. For positive $m$, this gives $-4{\pi }^2{\sum }_{r\ge 1}r{q}^{mr}$. Negative $m$ gives the same value, by replacing $n$ with $-n$ in its inner sum. The $m=0$ row is ${\sum }_{n\ne 0}{n}^{-2}={\pi }^2/3$, by the Basel problem. The resulting series in $m,r$ does have absolute convergence, locally uniformly in $z$, so collecting the coefficient at $q^{\ell }$ is legitimate:The coefficient is the sum-of-divisors function, since $r$ runs over the positive divisors of $\ell$. ThusThere is no conflict with vanishing of weight-two level-one modular forms: the Eisenstein series of weight two has an anomalous transformation term, so it is not a weight-two modular form.

Write the function from the preceding part as $F_m$. Its expansion at zero isHence the residue theorem givesChoose $R=R_m=(m+\frac{1}{2})\pi$. This circle stays a distance at least $\pi /2$ from every pole of $f$, so $f$ is bounded uniformly on it. MoreoverThus ${\max }_{C_{R_m}}|F_m|=O(\log m)$, and the ML inequality yieldsLetting $m\to \infty$ solves the Basel problem: