Сheck сonvergence $\sum _{n=2}^{\infty } \frac{\sqrt[n]{n}}{\ln^2{n}}$

How to check сonvergence of the following?

$$\sum _{n=2}^{\infty } \frac{\sqrt[n]{n}}{\ln^2{n}}$$

My attempt was that I took exponential by $a_{n} = \frac{\sqrt[n]{n}}{\ln^2{n}}$ : $e^{(\frac{\ln(n)}{n} - 2\ln{\ln{n}})}$ ~ $e^{(-2\ln{\ln{n}})}$ = $\frac{1}{\ln^2{n}}$.

Thus $$b_{n} = \frac{1}{\ln^2{n}}$$ $\sum _{n=2}^{\infty } b_{n}$ is divergent.

Is this correct solution? I would be very happy if you show other elegant solutions.

You see that for $n\geq 2$. As $n^{\frac{1}{n}}\geq 1$ we have $\displaystyle\frac{n^{\frac{1}{n}}}{\ln^{2}(n)}\geq \frac{1}{\ln^{2}(n)}$.

and the series $\displaystyle\sum_{n=2}^{\infty} \frac{1}{\ln^{2}(n)}$ diverges by Cauchy Condensation test. Hence By direct comparison , we have the original series is divergent.