{a_{r}} is an infinite sequence of real numbers. Let b_{n} = 1/n ∑_{1}^{n} exp(i a_{r}). Prove that b_{1}, b_{2}, b_{3}, b_{4}, ... converges to k iff b_{1}, b_{4}, b_{9}, b_{16}, ... 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 = n^{2} and N' = n^{2} + 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/n^{2} = 2/n. Also |(1/N - 1/m) ∑ c_{r} | ≤ (1/N - 1/m) ∑ |c_{r}| < (1/n^{2} - 1/(n+1)^{2}) (n+1)^{2} = (2n+1)/n^{2} < 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.

© John Scholes

jscholes@kalva.demon.co.uk

25 Jan 2002