Convergence of $a_n=\sum\limits_{m=2}^{n} \frac{(-1)^mm}{(\ln(m))^m}$ and $b_n = \sum\limits_{m=2}^{n} \frac{1}{(\ln(m))^m}$

Question

Let $a_1=b_1=0$ and for each $n\geq 2$, let $a_n$ and $b_n$ be real numbers given by
$a_n=\sum\limits_{m=2}^{n} \frac{(-1)^mm}{(\ln(m))^m}$ and $b_n = \sum\limits_{m=2}^{n} \frac{1}{(\ln(m))^m}$, then what can be said about the convergence of sequences $(a_n)$ and $(b_n)$?

If I am not wrong, is the question only asking whether the series

$\sum\limits_{m=2}^{\infty} \frac{(-1)^mm}{(\ln(m))^m}$ and $ \sum\limits_{m=2}^{\infty} \frac{1}{(\ln(m))^m}$ converge or diverge?

In which case they both converge by root test. What is the correct response to this?

Answer 1

Using Cauchy's n-th root test $a_n$ converges absolutely. Similarly $b_n$ converges using $n-th root test.