All the complex numbers z_{n} are non-zero and |z_{m} - z_{n}| > 1 (for any m ≠ n). Show that ∑ 1/z_{n}^{3} converges.

**Solution**

We show that it is absolutely convergent. Take a disk radius 1/2 centred on each z_{n}. The disks must be disjoint since |z_{m} - z_{n}| > 1 for distinct m, n. Now consider how many z_{n} can lie in the annulus N ≤ |z| < N+1. If z_{n} lies in the annulus, then at least half the corresponding disk must also, so if M is the number of z_{n}, then M π/8 < π ( (N+1)^{2} - N^{2}) = (2N+1) π, and the sum of 1/|z_{n}|^{3} for these z_{n} is at most M/N^{3} = 8(2N + 1)/N^{3}. But ∑ 8(2N + 1)/N^{3} converges, since ∑ 1/N^{2} converges.

© John Scholes

jscholes@kalva.demon.co.uk

11 Mar 2002