Proving the convergence of $\sum_{n=2}^{\infty} \frac{1}{n (\ln n)^2}$

Solution 1:

In the limit comparison test, $\sum_n f(n)$ and $\sum_n g(n)$ both will behave the same iff $$\lim_{n \to \infty} \dfrac{f(n)}{g(n)} = c \in (0,\infty)$$ In your case, the limit is $0$ and hence you cannot conclude anything.

Solution 2:

You can use integral test for convergence. Since the integral $$\int_2^\infty\dfrac{1}{x\log^2 x}dx$$ converges, the series $\displaystyle{\sum_{n=2}^{\infty} \frac{1}{n (\ln n)^2}}$ converges.

Solution 3:

Expanding on the accepted answer, you should use the Cauchy condensation test: For a positive non-increasing sequence $f(n)$, the series $\sum_{n=0}^\infty f(n)$ converges if and only if the series $\sum_{n=0}^\infty2^nf(2^n)$ does.
In your case, $f(n)=\frac{1}{n\ln^2n}$, so $$2^nf(2^n)=2^n\frac{1}{2^n\ln^22^n}=\frac{1}{n^2\ln^22}=\frac{1}{\ln^22}\frac{1}{n^2}$$ which clearly converges.