Showing convergence or divergence of a sequence I need to determine if the series with $n$th term $\ln(n)e^{-\sqrt n}$ converges or diverges. I've tried numerous identities for $ln(x)$ and $e^{x}$ and various convergence tests but I'm still very stuck. 

For variety, lets use the Cauchy condensation test as an alternative to the integral test for $S=\sum_{n=1}^{\infty}\ln(n)e^{-\sqrt n}$. Then the convergence of this series is identical to the convergence of the condensed series
$$
T=\sum_{k=0}^{\infty}2^k\ln(2^k)e^{-\sqrt {2^k}}
$$
Now use for $x>0$ the very coarse estimate
$$
e^{-x}=\frac1{1+x+\frac{x^2}2+\frac{x^3}6+\frac{x^4}{4!}+...}\le\frac{24}{x^4}\tag{*}
$$ 
to get the upper estimate
$$
T\le24\ln2\,\sum_{k=0}^{\infty}\frac{k}{2^k}=48\ln(2).
$$
So $T$ converges and in consequence also $S$ converges.

Or to stay even more elementary, use asymptotic estimates and note that 
$$
\lim_{n\to\infty}\ln(n)\,n^2\,e^{-\sqrt{n}}=0,
$$ 
by using for instance that $e^{-\sqrt{n}}\le\frac1{1+\frac{n^4}{8!}}$, also per (*),
so that for some $n_0\in\Bbb N$ and all $n\ge n_0$ one gets the convergent majorant 
$$
\ln(n)\,e^{-\sqrt{n}}\le \frac1{n^2}.
$$
(Seen too late: This last is nearly the same idea as in the answer of E-Theory.)
