 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2017/ia/paper-1/11f/a/solution

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

For each

k \geq 0

, monotonicity gives the dyadic-block bounds

2^{k} x_{2^{k + 1}} \leq \sum_{n = 2^{k}}^{2^{k + 1} - 1} x_{n} \leq 2^{k} x_{2^{k}} .

Summing the upper bounds proves convergence of

\sum x_{n}

whenever

\sum 2^{k} x_{2^{k}}

converges. Summing the lower bounds gives, up to the first term,

\sum_{k \geq 0} 2^{k} x_{2^{k + 1}} = \frac{1}{2} \sum_{j \geq 1} 2^{j} x_{2^{j}},

so convergence of

\sum x_{n}

forces convergence of the condensed series. This is the Cauchy condensation test:

n \geq 1 \sum x_{n} converges ⟺ k \geq 0 \sum 2^{k} x_{2^{k}} converges .

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