The sequence an is monotonic and ∑ an converges. Show that ∑ n(an - an+1) converges.
Solution
The sum of the first n terms is (a1 - a2) + 2(a2 - a3) + ... + n(an - an+1) = a1 + a2 + a3 + ... + an - n an+1. We are given that ∑ an converges, so it is sufficient to show that the sequence n an+1 converges to zero.
But since ∑ an converges, |an+1 + an+2 + ... + a2n| is arbitrarily small for n sufficiently large. Since an is monotonic, this implies that n an+1 is arbitrarily small for n sufficiently large.
© John Scholes
jscholes@kalva.demon.co.uk
5 Mar 2002