The sequence a_{1} + 2a_{2}, a_{2} + 2a_{3}, a_{3} + 2a_{4}, ... converges. Prove that the sequence a_{1}, a_{2}, a_{3}, ... also converges.

**Solution**

Note that the result is not true for a_{i} + a_{2}, a_{2} + a_{3}, ... or for 2a_{1} + a_{2}, 2a_{2} + a_{3}, ... . In the first case, we could have 1, -1, 1, -1, 1, ... . In the second case, we could have 1, -2, 4, -8, 16, -32, ... .

Suppose a_{n} + 2a_{n+1} converges to 3k. We show that a_{n}converges to k.

Given any ε > 0, take N so that a_{n} + 2a_{n+1} is within ε of 3k for all n >= N. Take a positive integer M such that a_{N} is within (2^{M} + 1)ε of k.

Then a_{N+1} is within ( (2^{M} + 1)ε + ε)/2 = (2^{M-1} + 1)ε of (3k - k)/2 = k. By a trivial induction a^{N+M} is within 2ε of k. Then a^{N+M+1} is within (2ε + ε)/2, and hence within 2ε, of k. So by a trivial induction, a_{n} is within 2ε of k for all n > N + M.

*Comment. This is identical to 1950/B5 (1).*

© John Scholes

jscholes@kalva.demon.co.uk

14 Jan 2002