= Solution
Write the function from the preceding part as $F_m$. Its expansion at zero is
$$
F_m(z)=-\frac12+z\left(\frac16-
\frac1{\pi^2}\sum_{k=1}^m\frac1{k^2}\right)+O(z^3).
$$
Hence the <residue theorem> gives
$$
\frac1{2\pi i}\oint_{C_R}\frac{F_m(z)}{z^2}\,dz
=F_m'(0)=\frac16-\frac1{\pi^2}\sum_{k=1}^m\frac1{k^2}.
$$
Choose $R=R_m=(m+\tfrac12)\pi$. This circle stays a distance at least $\pi/2$ from every pole of $f$, so $f$ is bounded uniformly on it. Moreover
$$
\sum_{k=1}^m\left|\frac z{z^2+\pi^2k^2}\right|
\leq\sum_{k=1}^m\frac{R_m}{R_m^2-\pi^2k^2}
=O(\log m).
$$
Thus $\max_{C_{R_m}}|F_m|=O(\log m)$, and the <ML inequality> yields
$$
\left|\oint_{C_{R_m}}\frac{F_m(z)}{z^2}\,dz\right|
=O\left(\frac{\log m}{m}\right)\longrightarrow0.
$$
Letting $m\to\infty$ solves the <Basel problem>:
$$
\boxed{\sum_{k=1}^{\infty}\frac1{k^2}=\frac{\pi^2}{6}.}
$$
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