Find a constant k such that for any positive ai, ∑1∞ n/(a1 + a2 + ... + an) ≤ k ∑1∞1/an.
Let us write sn = a1 + a2 + ... + an and bn= n/sn. If ∑ 1/an diverges, then there is nothing to prove. So assume it converges. Hence an must diverge, so there are only finitely many values less than any given K, so we can arrange the terms in increasing order c1 ≤ c2 ≤ c3 ... . Since all the terms are positive ∑ 1/an is absolutely convergent and equal to ∑ 1/cn. Also we have sn ≥ c1 + c2 + ... + cn. The trick is to notice that half the ci are at least cn/2. It is convenient to consider separately n = 2m and n = 2m+1. So s2m ≥ m cm. Hence b2m < 2/cm. Similarly, s2m-1 > m cm, so b2m-1 < 2/cm. Hence ∑12m bi ≤ 4 ∑1m 1/ci, which gives the required result with k = 4.
25th Putnam 1964
© John Scholes
5 Feb 2002