For which real p does the series sum over n >= 2 of 1/(n (log n)^p) converge? Prove the boundary case fails, and say why a direct comparison against 1/n^s for any s > 1 cannot settle it.

For which real p does the series sum over n >= 2 of 1/(n (log n)^p) converge? Prove the boundary case fails, and say why a direct comparison against 1/n^s for any s > 1 cannot settle it.

Approach: Group the terms into dyadic blocks between 2^k and 2^{k+1}, which turns the log into a linear factor, or equivalently apply the integral test with the substitution u = log x.

The series converges exactly for p > 1. Cauchy condensation replaces the sum by sum 2^k * 1/(2^k (log 2^k)^p) = sum 1/(k log 2)^p, which is a p-series in k and converges precisely when p > 1. The integral test gives the same answer directly: with u = log x, the integral of 1/(x (log x)^p) dx becomes the integral of u^{-p} du, finite at infinity only for p > 1. At the boundary p = 1 the antiderivative is log log x, which grows without bound, so the series diverges. A comparison test against 1/n^s cannot decide the question because for any s > 1 the ratio (1/(n (log n)^p)) / (1/n^s) = n^{s-1}/(log n)^p tends to infinity, so the terms are eventually larger than the convergent comparison, while against 1/n they are eventually smaller than the divergent one. Every power of n is either too coarse above or too coarse below, which is why the log scale itself has to be the comparison.

Follow-up: Does sum 1/(n log n (log log n)^p) follow the same rule, and what does the pattern say about how slowly a divergent series can diverge?

Key concepts: cauchy condensation test, integral test, comparison test, divergence at the boundary.