25th Putnam 1964

Find a constant k such that for any positive a_i, ∑₁^∞ n/(a₁ + a₂ + ... + a_n) ≤ k ∑₁^∞1/a_n.

Solution

Let us write s_n = a₁ + a₂ + ... + 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 c₁ ≤ c₂ ≤ c₃ ... . Since all the terms are positive ∑ 1/a_n is absolutely convergent and equal to ∑ 1/c_n. Also we have s_n ≥ c₁ + c₂ + ... + 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 < 2/c_m. Similarly, s_2m-1 > 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