11th Putnam 1951

Show that if ∑ a_n converges, then so does ∑ a_n/n.

Solution

Put s_n = a₁ + a₂ + ... + a_n. Then s_n - s_n-1 = a_n, and so a₁/1 + a₂/2 + a₃/3 + ... + a_n/n = s₁(1/1 - 1/2) + s₂(1/2 - 1/3) + ... + s_n(1/n - 1/n+1) + s_n/n+1. Now s₁(1/1 - 1/2) + s₂(1/2 - 1/3) + ... + s_n(1/n - 1/n+1)+ ... is absolutely convergent, because the sequence |s_n| is bounded, say by B. Hence |s₁(1/1 - 1/2)| + |s₂(1/2 - 1/3)| + ... + |s_n(1/n - 1/n+1)| <= B( (1/1 - 1/2) + (1/2 - 1/3) + ... + (1/n - 1/n+1) ) = B(1 - 1/n+1) < B. Obviously s_n/n+1 tends to zero, so ∑ a_n/n is absolutely convergent and hence convergent.

Comment. This is Abel's summation formula.

11th Putnam 1951

© John Scholes
jscholes@kalva.demon.co.uk
5 Mar 2002