24th Putnam 1963

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Problem B5

The series ∑ an of non-negative terms converges and ai ≤ 100an for i = n, n + 1, n + 2, ... , 2n. Show that limn→∞ nan = 0.

 

Solution

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
jscholes@kalva.demon.co.uk
5 Feb 2002