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

I am attempting to prove if the below expression converges or diverges:

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

I decided to try using the limit convergence test. So then, I set $f(n) = \frac{1}{n (\ln n)^2}$ and $g(n) = \frac{1}{n}$ and did the following:

$$
\lim_{n \to \infty} \frac{f'(n)}{g'(n)} =
\lim_{n \to \infty} \frac{-\frac{\ln(n) + 2}{n^2 (\ln n)^3}}{-\frac{1}{n^2}} =
\lim_{n \to \infty} \frac{\ln(n) + 2}{n^2 (\ln n)^3} \cdot n^2 =
\lim_{n \to \infty} \frac{1}{(\ln n)^2} + \frac{2}{(\ln n)^3} = 0

$$

I know that $\sum \frac{1}{n}$ diverges due to properties of harmonic series, and so concluded that my first expression $\frac{1}{n (\ln n)^2}$ also must diverge.

However, according to Wolfram Alpha, and as suggested by this math.SE question, the expression actually converges.

What did I do wrong?

Answer

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.