Find the values of p for which the series converges. ∑n=1 to ∞ for (ln(n)/(n^(2p)))

Question

Answer in interval notation

Answer 1

a_n = lnn/n^2p = lnn/n·1/n^2p-1; 1/n^2p < a_n < 1/n^2p-1, starting from n = 3.

So, p-series property: 2p > 1 and 2p - 1 > 1; p > 1/2 and p > 1. Answer p > 1