The series ∑ a_{n} of non-negative terms converges and a_{i} ≤ 100a_{n} for i = n, n + 1, n + 2, ... , 2n. Show that lim_{n→∞} na_{n} = 0.

**Solution**

We need to invert the inequality given. We are given a collection of a_{i} which are less than a fixed a_{n}. We want to fix a_{n} and find a collection of a_{j} such that a_{n} ≤ 100 a_{j}.

Evidently, a_{2n} ≤ 100 a_{2n-1}, a_{2n} ≤ 100 a_{2n-2}, a_{2n} ≤ 100 a_{2n-3}, ... , a_{2n} ≤ 100 a_{n}. Adding and multiplying by two, 2n a_{2n} ≤ 200 (a_{n} + a_{n+1} + ... + a_{2n-1}). But ∑ a_{n} converges, so (a_{n} + a_{n+1} + ... + a_{2n-1}) < ε/200 for all sufficiently large n, and hence 2n a_{2n} < ε for sufficiently large n. Similarly for (2n+1) a^{2n+1}. So n a_{n} tends to zero.

© John Scholes

jscholes@kalva.demon.co.uk

5 Feb 2002