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. 

Assuming you are referring to the sequence $\left\langle \ln(n)e^{-\sqrt{n}} \right\rangle_{n=1}^\infty$, and not the series $\left \langle \sum_{k=1}^n \ln(k)e^{-\sqrt{k}} \right \rangle_{n=1}^\infty$, then using L'Hôpital's rule, we see that:
$$
\lim_{n \to \infty} \frac{\ln(n)}{e^{\sqrt{n}}} = \lim_{n \to \infty} \frac{\frac 1n}{e^{\sqrt{n}} \left( \frac{1}{2\sqrt{n}} \right)} = \lim_{n \to \infty} \frac{2}{\sqrt{n}e^{\sqrt{n}}} = 2\lim_{n \to \infty}n^{-1/2} \cdot \lim_{n \to \infty} e^{-\sqrt{n}} = 2 \cdot 0 \cdot 0 = 0
$$
Therefore, the sequence converges to $0$.
