The series ∑ an of non-negative terms converges and ai ≤ 100an for i = n, n + 1, n + 2, ... , 2n. Show that limn→∞ nan = 0.
We need to invert the inequality given. We are given a collection of ai which are less than a fixed an. We want to fix an and find a collection of aj such that an ≤ 100 aj.
Evidently, a2n ≤ 100 a2n-1, a2n ≤ 100 a2n-2, a2n ≤ 100 a2n-3, ... , a2n ≤ 100 an. Adding and multiplying by two, 2n a2n ≤ 200 (an + an+1 + ... + a2n-1). But ∑ an converges, so (an + an+1 + ... + a2n-1) < ε/200 for all sufficiently large n, and hence 2n a2n < ε for sufficiently large n. Similarly for (2n+1) a2n+1. So n an tends to zero.
24th Putnam 1963
© John Scholes
5 Feb 2002