Infinite series $sum _{n=2}^{infty } frac{1}{n log (n)}$

Recently, I encountered a problem about infinite series.
So my question is how to know whether the infinite series $\sum _{n=2}^{\infty } \frac{1}{n \log (n)}$ is convergent?

Answer

To see whether $\sum_2^\infty 1/(n \log n)$ converges, we can use the integral test. This series converges if and only if this integral does:

$$
\int_2^\infty \frac{1}{x \log x} dx = \left[\log(\log x)\right]_2^\infty
$$
and in fact the integral diverges.

This is part of a family of examples worth remembering. Note that
$$
d/dx \log(\log(\log x)) = d/dx \log(\log x) \cdot \frac{1}{\log (\log x)} = \frac{1}{x \log x \log(\log x)}
$$
and $\log (\log (\log x)) \to \infty$ as $x \to \infty$ hence $\sum \frac{1}{n \log n \log (\log n)}$ diverges as well. Similarly, by induction we can put as many iterated $\log$s in the denominator as we want (i.e. $\sum \frac{1}{n \log n \log(\log n) \ldots \log (\ldots (\log n) \ldots )}$ where the $i$th log is iterated $i$ times), and it will still diverge. However, as you should check, $\sum \frac{1}{x \log^2x}$ converges, and in fact (again by induction) if you square any of the iterated logs in $\sum \frac{1}{n \log n \log(\log n) \ldots \log (\ldots (\log n) \ldots )}$ the sum will converge.