Putnam 1963/B5 solution

24th Putnam 1963

Problem B5

The series ∑ a_n of non-negative terms converges and a_i ≤ 100a_n for i = n, n + 1, n + 2, ... , 2n. Show that lim_n→∞ na_n = 0.

Solution

We need to invert the inequality given. We are given a collection of a_i which are less than a fixed a_n. We want to fix a_n and find a collection of a_j such that a_n ≤ 100 a_j.

Evidently, a_2n ≤ 100 a_2n-1, a_2n ≤ 100 a_2n-2, a_2n ≤ 100 a_2n-3, ... , a_2n ≤ 100 a_n. Adding and multiplying by two, 2n a_2n ≤ 200 (a_n + a_n+1 + ... + a_2n-1). But ∑ a_n converges, so (a_n + a_n+1 + ... + a_2n-1) < ε/200 for all sufficiently large n, and hence 2n a_2n < ε for sufficiently large n. Similarly for (2n+1) a²ⁿ⁺¹. So n a_n tends to zero.

24th Putnam 1963