Testing Convergence of $\sum \sqrt{\ln{n}\cdot e^{-\sqrt{n}}}$ What test should i apply for testing the convergence/divergence of $$\sum_{n=1}^{\infty} \sqrt{\ln{n}\cdot e^{-\sqrt{n}}}$$
Help with hints will be appreciated.
Thanks

The $n$-th term is equal to
$$\frac{\sqrt{\log n}}{e^{\sqrt{n}/2}}.$$
The intuition is that the bottom grows quite fast, while the top does not grow fast at all.
In particular, after a while the top is $\lt n$. 
If we can show, for example, that after a while $e^{\sqrt{n}/2}\gt n^3$, then by comparison with $\sum \frac{1}{n^2}$ we will be finished.
So is it true that in the long run $e^{\sqrt{n}/2}\gt n^3$? Equivalently, is it true that in the long run $\sqrt{n}/2\gt 3\log n$? Sure, in fact $\lim_{n\to\infty}\dfrac{\log n}{\sqrt{n}}=0$, by L'Hospital's Rule, and in other ways. 
Remark: A named test that works well here is the Cauchy Condensation Test. I believe that a more "hands on" confrontation with the decay rate of the $n$-th term is more informative. 
