Let B_{n}(x) = 1^{x} + 2^{x} + ... + n^{x} and let f(n) = B_{n}(log_{n}2) / (n log_{2}n)^{2}. Does f(2) + f(3) + f(4) + ... converge?

**Solution**

Answer: yes.

This looks much worse than it is. A crude estimate for B_{n}(log_{n}2) is sufficient. We know that ∑ 1/n only just fails to converge. So it is likely that ∑ 1/n 1/log^{2}n will converge. In that case we need only prove that B_{n}(log_{n}2) ≤ O(n). But it has n terms, the biggest of which is 2. So certainly B_{n}(log_{n}2) < 2n.

We can easily check that ∑ 1/n 1/log^{2}n converges by the integral test (the indefinite integral is -1/log n).

© John Scholes

jscholes@kalva.demon.co.uk

16 Jan 2001