 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ia/paper-2/10f/v/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

Part (iii) and the given harmonic-sum asymptotic show that

\frac{E T _{n}}{n l o g n} = \frac{H _{n}}{l o g n} ⟶ 1.

For any fixed

ε > 0

, this deterministic ratio lies within

ε /2

1

for all sufficiently large

n

. The Chebyshev inequality and part (iv) then give

P (\frac{T _{n}}{n lo g n} - 1 > ε) \leq P (∣ T_{n} - E T_{n} ∣ > \frac{ε}{2} n lo g n) \leq \frac{4 var ( T _{n} )}{ε ^{2} n ^{2} ( lo g n ) ^{2}} \leq \frac{4 C}{ε ^{2} ( lo g n ) ^{2}} ⟶ 0.

Thus

\frac{T _{n}}{n lo g n} ⟶ 1 in probability.

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