26th Putnam 1965

{a_r} is an infinite sequence of real numbers. Let b_n = 1/n ∑₁ⁿ exp(i a_r). Prove that b₁, b₂, b₃, b₄, ... converges to k iff b₁, b₄, b₉, b₁₆, ... converges to k.

Solution

If a sequence converges to k, then any subsequence also converges to k.

So suppose that the square terms converge to k. Let c_r = exp(i a_r). Take two consecutive squares N = n² and N' = n² + 2n + 1 and m between them. Then |b_N - b_m| ≤ |b_N - 1/N ∑ c_r| + |(1/N - 1/m) ∑ c_r |, where the sums are over m terms. But |b_N - 1/N ∑ c_r| ≤ 1/N ∑_N+1^m |c_r| = (m - N)/N ≤ 2n/n² = 2/n. Also |(1/N - 1/m) ∑ c_r | ≤ (1/N - 1/m) ∑ |c_r| < (1/n² - 1/(n+1)²) (n+1)² = (2n+1)/n² < 3/n. So the difference |b_m - b_N| is arbitrarily small for n sufficiently large. Thus if we take m sufficiently large, then |b_m - b_N| < ε and |b_N - k| < ε. So |b_m - k| < 2ε and the sequence converges.

26th Putnam 1965

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