The sequence a1 + 2a2, a2 + 2a3, a3 + 2a4, ... converges. Prove that the sequence a1, a2, a3, ... also converges.
Note that the result is not true for ai + a2, a2 + a3, ... or for 2a1 + a2, 2a2 + a3, ... . 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 an + 2an+1 converges to 3k. We show that anconverges to k.
Given any ε > 0, take N so that an + 2an+1 is within ε of 3k for all n >= N. Take a positive integer M such that aN is within (2M + 1)ε of k.
Then aN+1 is within ( (2M + 1)ε + ε)/2 = (2M-1 + 1)ε of (3k - k)/2 = k. By a trivial induction aN+M is within 2ε of k. Then aN+M+1 is within (2ε + ε)/2, and hence within 2ε, of k. So by a trivial induction, an is within 2ε of k for all n > N + M.
Comment. This is identical to 1950/B5 (1).
30th Putnam 1969
© John Scholes
14 Jan 2002