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
 A: Hint:  Lets use the comparison test.  To understand what we should compare to, lets look at the logarithm of the terms in the series, as this will be easier to get a grasp on.  Notice that $$\log\left(\sqrt{e^{-\sqrt{n}}\log n}\right)=-\frac{1}{2}\sqrt{n}+\frac{1}{2}\log\log n\leq -2\log n$$ for sufficiently large $n$.  The last inequality follows since $2\log n+\frac{1}{2}\log \log n \leq \frac{1}{2}\sqrt{n}$ for large $n$, as the square root function grows much quicker than the logarithm. This tells us that $$\sqrt{e^{-\sqrt{n}}\log n}\leq\frac{1}{n^{2}}$$ for sufficiently large $n$.
Idea: The function $e^{n^\alpha}$ for $0<\alpha<1$ grows slower than any exponential function, $e^{Cn}$ but it grows faster than any power $n^\beta$.
A: 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. 
