 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ib/paper-2/13a/d/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

Write the function from the preceding part as

F_{m}

. Its expansion at zero is

F_{m} (z) = - \frac{1}{2} + z (\frac{1}{6} - \frac{1}{π ^{2}} \sum_{k = 1}^{m} \frac{1}{k ^{2}}) + O (z^{3}) .

Hence the residue theorem gives

\frac{1}{2 πi} \oint_{C_{R}} \frac{F _{m} ( z )}{z ^{2}} d z = F_{m}^{'} (0) = \frac{1}{6} - \frac{1}{π ^{2}} \sum_{k = 1}^{m} \frac{1}{k ^{2}} .

Choose

R = R_{m} = (m + \frac{1}{2}) π

. This circle stays a distance at least

π /2

from every pole of

f

, so

f

is bounded uniformly on it. Moreover

\sum_{k = 1}^{m} \frac{z}{z ^{2} + π ^{2} k ^{2}} \leq \sum_{k = 1}^{m} \frac{R _{m}}{R _{m}^{2} - π ^{2} k ^{2}} = O (lo g m) .

Thus

max_{C_{R_{m}}} ∣ F_{m} ∣ = O (lo g m)

, and the ML inequality yields

\oint_{C_{R_{m}}} \frac{F _{m} ( z )}{z ^{2}} d z = O (\frac{l o g m}{m}) ⟶ 0.

Letting

m \to \infty

k = 1 \sum \infty \frac{1}{k ^{2}} = \frac{π ^{2}}{6} .

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