 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ia/paper-1/3f/c/solution

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

Another application of the Cauchy-Schwarz inequality gives

\sum_{n = 1}^{N} a_{n} n^{- p} \leq (\sum_{n = 1}^{N} a_{n})^{1/2} (\sum_{n = 1}^{N} n^{- 2 p})^{1/2} .

When

p > 1/2

, the second factor is bounded because

2 p > 1

, so the series converges.

At the endpoint, take

a_{1} = 1

and, for

n \geq 2

a_{n} = \frac{1}{n ( l o g n ) ^{2}} .

Then

\sum a_{n}

converges, while

a_{n} n^{- 1/2} = \frac{1}{n l o g n},

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