30th Putnam 1969

The sequence a₁ + 2a₂, a₂ + 2a₃, a₃ + 2a₄, ... converges. Prove that the sequence a₁, a₂, a₃, ... also converges.

Solution

Note that the result is not true for a_i + 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 a_n + 2a_n+1 converges to 3k. We show that a_nconverges 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).

30th Putnam 1969

© John Scholes
jscholes@kalva.demon.co.uk
14 Jan 2002