Convergence of a ln-series

Question

I have been wondering about whether the following sequence converges or not: $$ \sum_{n=2}^\infty \frac{1}{(\ln(n))^{2p} n^2}, $$where $p \leq 1$. I believe that for $p \leq 0$ we can show that this series can be bounded above by $\sum_{n\geq2}\frac{1}{n^4}$, which obviously converges. I can't seem to find a way of showing what happens for $0 < p \leq 1$. I believe that for these values this series also converges, but I am not sure on how to show this.

Answer 1

In the case that $0 \leq p \leq 1$: let's look at the same series, but summed from $n \geq 3$. Obviously, if this series converges, then so does the original one. Let's define $f(n) = \frac{1}{(\ln n)^{2p} n^2}$. Since $n \geq 3$, $\ln n \geq 1$. Now, let's look at two cases:

1. $p \leq 0.5$ or equivalently, $2p \leq 1$. Then $(\ln n)^{2p} \geq 1$, and therefore $f(n) \leq \frac{1}{n^2}$, whose series converges.
2. $p > 0.5$ or equivalently, $2p > 1$. Then $(\ln n)^{2p} \geq \ln n$, and therefore $f(n) \leq \frac{1}{n^2 \ln n}$, whose series also converges.

So this is convergent for all $0 \leq p \leq 1$.