$$\displaystyle\sum_{n=2}^\infty\frac{\exp(n)}{\exp(n\sqrt[n]{n})\ln^{2}n}$$
Tell, if the series converges.
Please verify my answer.
$$\displaystyle \sum_{n=2}^\infty\dfrac{\exp(n)}{\exp(n\sqrt[n]{n})\ln^{2}n}=\sum_{n=2}^\infty\dfrac{\exp(n)}{\exp(n)\frac{1}{n}\exp(n)\ln^{2}n}=\sum_{n=2}^\infty\dfrac{n}{\exp(n)\ln^{2}n}$$
Comparison test
$$\displaystyle \lim_{n\to \infty}\frac{n+1}{\exp(n+1)\ln^{2}(n+1)}\frac{\exp(n)\ln^{2}n}{n}=\underset{n\rightarrow\infty}{\lim}\frac{1}{e}\frac{n+1}{n}\frac{\ln^{2}n}{\ln^{2}(n+1)}=\frac{1}{e}$$
The series converges.
