25th Putnam 1964 Problem A5 Find a constant k such that for any positive a i , ∑ 1 ∞ n/(a 1 + a 2 + ... + a n ) ≤ k ∑ 1 ∞ 1/a n .
Let us write s n = a 1 + a 2 + ... + a n and b n = n/s n . If ∑ 1/a n diverges, then there is nothing to prove. So assume it converges. Hence a n must diverge, so there are only finitely many values less than any given K, so we can arrange the terms in increasing order c 1 ≤ c 2 ≤ c 3 ... . Since all the terms are positive ∑ 1/a n is absolutely convergent and equal to ∑ 1/c n . Also we have s n ≥ c 1 + c 2 + ... + c n . The trick is to notice that half the c i are at least c n/2 . It is convenient to consider separately n = 2m and n = 2m+1. So s 2m ≥ m c m . Hence b 2m  m c m , so b 2m-1 < 2/c m . Hence ∑ 1 2m b i ≤ 4 ∑ 1 m 1/c i , which gives the required result with k = 4. 25th Putnam 1964 © John Scholes jscholes@kalva.demon.co.uk 5 Feb 2002