 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ia/paper-1/11e/b/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

a_{n} a_{n + 1} \leq \frac{a _{n} + a _{n + 1}}{2} .

Thus convergence of

\sum_{n} a_{n}

implies convergence of

\sum_{n} a_{n} a_{n + 1}

The converse is false. Define

a_{2 n} = \frac{1}{n}, a_{2 n - 1} = \frac{1}{n ^{3}} .

The even terms make

\sum_{n} a_{n}

diverge, while

a_{2 n - 1} a_{2 n} = \frac{1}{n ^{2}}, a_{2 n} a_{2 n + 1} = \frac{1}{n ( n + 1 ) ^{3}} \leq \frac{1}{n ^{2}},

so the neighboring geometric-mean series converges.

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