Show that if ∑ a_{n} converges, then so does ∑ a_{n}/n.

**Solution**

Put s_{n} = a_{1} + a_{2} + ... + a_{n}. Then s_{n} - s_{n-1} = a_{n}, and so a_{1}/1 + a_{2}/2 + a_{3}/3 + ... + a_{n}/n = s_{1}(1/1 - 1/2) + s_{2}(1/2 - 1/3) + ... + s_{n}(1/n - 1/n+1) + s_{n}/n+1. Now s_{1}(1/1 - 1/2) + s_{2}(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 |s_{1}(1/1 - 1/2)| + |s_{2}(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.*

© John Scholes

jscholes@kalva.demon.co.uk

5 Mar 2002