{title2=$G_2(z)=\sum_m\sum_n'(mz+n)^{-2}$}

In the displayed iterated order, with the $n$ sum evaluated first, $G_2(z)=(\pi^2/3)E_2(z)$. The <cosecant partial-fraction identity> $\sum_n(w+n)^{-2}=\pi^2\csc^2(\pi w)$ gives the positive-$m$ contribution $-4\pi^2\sum_{r\geq1}r q^{mr}$; negative $m$ gives the same contribution. Including the $m=0$ term yields $\pi^2/3-8\pi^2\sum_{n\geq1}\sigma_1(n)q^n$. The full lattice series does not have <absolute convergence>, so this order must not be replaced by arbitrary rearrangement.

Iterated Eisenstein summation in weight two

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> is
$$
\sum_{n\in\mathbb Z}\frac1{(w+n)^2}=\pi^2\csc^2(\pi w).
$$
For 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 $\operatorname{Im}w>0$, the geometric-series expression $\cot(\pi w)=-i(1+2\sum_{r\geq1}e^{2\pi irw})$, differentiated termwise, gives the <cotangent partial-fraction Fourier kernel>
$$
\pi^2\csc^2(\pi w)=-4\pi^2\sum_{r\geq1}r e^{2\pi irw}.
$$

Put $q=e^{2\pi iz}$ with $z$ in the <complex upper half-plane>. For positive $m$, this gives $-4\pi^2\sum_{r\geq1}r q^{mr}$. Negative $m$ gives the same value, by replacing $n$ with $-n$ in its inner sum. The $m=0$ row is $\sum_{n\ne0}n^{-2}=\pi^2/3$, 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:
$$
G_2(z)=\frac{\pi^2}{3}-8\pi^2\sum_{m,r\geq1}r q^{mr}
=\frac{\pi^2}{3}-8\pi^2\sum_{\ell\geq1}\sigma_1(\ell)q^\ell.
$$
The coefficient is the <sum-of-divisors function>, since $r$ runs over the positive divisors of $\ell$. Thus
$$
\boxed{G_2(z)=\frac{\pi^2}{3}E_2(z).}
$$
There 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>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 137 / 3 / a / Solution

{title2=$\sum_{n\in\mathbb Z}(w+n)^{-2}=\pi^2\csc^2(\pi w)$}

For $w\notin\mathbb Z$, integrate $\pi\cot(\pi\zeta)/(\zeta-w)^2$ around large squares whose sides have half-integer coordinates. The <cotangent> is uniformly bounded on these sides and the integral tends to zero. The <residue theorem> gives residues $(n-w)^{-2}$ at the integers and $-\pi^2\csc^2(\pi w)$ at $w$, proving the identity. For $\operatorname{Im}w>0$, the geometric-series formula for the <cotangent> also gives $\pi^2\csc^2(\pi w)=-4\pi^2\sum_{r\geq1}r e^{2\pi irw}$.

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