Proving the divergence of $\sum_2^\infty{\frac{1}{n \log n}}$ using comparison test

Question

It is quite straightforward using the Cauchy Condensation test. But is there any way to solve this problem using some well known comparison test?

I cannot think of any way of my own. Any help/hint is appreciated.

Answer 1

The function $$f(x):= \frac{1}{x\log(x)}$$ is for $x > 2$ decreasing. Thus we get $$\sum_{n=2}^m \frac{1}{n \log(n)} \ge \int_{3}^{m} \frac{1}{x \log(x)} \, d x = \log \log m - \log \log (3)$$ The same argument can be used to find an upper bound. This gives the asymptotic formula $$\sum_{n=2}^m \frac{1}{n \log(n)} \sim \log \log m.$$