Is the series $ L = \sum_{i=2}^{\infty } \frac{1}{n \log(n)} $ convergent or divergent?

Question

Does $ L = \sum_{i=2}^{\infty } \frac{1}{n \log(n)} $ converge or diverge?

I established that: $$ L \le I = \int^{\infty}_{2} \frac{1}{n \log(n)} = \lim_{n \to \infty} [ \ln(\log(n)) - \ln(\log(2)) ], $$ and as $ \lim_{n \to \infty} \log(x) = \infty $, then $ L $ diverges.

But I'm not sure:

1. of the sense of the inequality,
2. about the conclusion.

Answer 1

What you want to use is the integral test: the series (with positive terms) $\;\sum_{n=2}^\infty\frac1{n\log n}$ converges (resp. diverges) if and only if the improper integral $\;\int_2^\infty\frac 1{x\log x}\,\mathrm dx$ converges (resp. diverges).

Now this integral is easy to calculate by substitution: setting $u=\log t$, we obtain $$\int_2^A \frac 1{x\log x}\,\mathrm dx=\int_{\log 2}^{\log A} \frac{\mathrm d u}u= \log\, \biggl|_{\log 2}^{\log A}=\log(\log A)-\log(\log 2)\xrightarrow[A\to\mkern1mu+\infty]{}+\infty.$$