Given series is convergent or divergent?

Question

$$\sum_{n=2}^{\infty} \frac{1}{(\ln(n))^2}$$

I thought I can compare it with $\frac{1}{n}$ but I couldn't prove it (and be satisfied with it). Therefore, how should I approach these type of series ? Thanks in advance..

Answer 1

You have $$ \lim_{n \to \infty}\frac{\ln^2 n}{n} = \lim_{n \to \infty}\frac{2 \ln n \cdot (1/n)}{1} = \lim_{n \to \infty}\frac{2\ln n}{n} = \lim_{n \to \infty}\frac{2/n}{1} = 0 $$ by using L'Hospital's rule twice. Hence, for large $n$, $\ln^2 n < n$ and now use comparison test.

Answer 2

For the series, you can use the integral test: since for $n\ge 2$, $\;x\ln x>\ln^2 x>0$, the improper integral $$\int _2^\infty\frac{\mathrm dx}{\ln^2x}\ge\int _2^\infty\frac{\mathrm dx}{x\ln x}=\ln (\ln x)\biggm |_2^{+\infty}=+\infty $$ diverges.

Added: This series is a particular case of the *Bertrand's series: $$\sum_{n=2}^\infty \frac 1{n^\alpha \ln^\beta n},$$ which are known to converge if and only if $\;(\alpha >1)$ or $(\alpha =1$ and $\beta >1)$.