Putnam 49/A3 solution

9th Putnam 1949

Problem A3

All the complex numbers z_n are non-zero and |z_m - z_n| > 1 (for any m ≠ n). Show that ∑ 1/z_n³ 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 - N²) = (2N+1) π, and the sum of 1/|z_n|³ for these z_n is at most M/N³ = 8(2N + 1)/N³. But ∑ 8(2N + 1)/N³ converges, since ∑ 1/N² converges.

9th Putnam 1949