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Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 3 / 26H / b / Solution

Here

P (∣ X_{n} ∣ > ε) = \frac{1}{n} ⟶ 0

for every

0 < ε < 1

, so

X_{n} \to 0

in probability. On the other hand, the events

A_{n} = {X_{n} = 1}

are independent and

\sum_{n = 1}^{\infty} P (A_{n}) = \sum_{n = 1}^{\infty} \frac{1}{n} = \infty.

The second Borel-Cantelli lemmas gives

P (X_{n} = 1 infinitely often) = 1.

Thus

X_{n}

cannot converge almost surely to zero. This independent rare-event counterexample to almost-sure convergence proves that convergence in probability need not imply almost-sure convergence of the full sequence.

Solved by gpt-5.6-sol high.

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