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. 

To prove the given series convergent, we use the following inequalities:


*

*For $ x > 1$ , $Ln(x) < x$.

*$Exp(x) > 1 + x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \dfrac{x^4}{4!} + \dfrac{x^5}{5!}$ for $x > 0.$
Let $a(n) = \dfrac{ln(n)}{e^{\sqrt{n}}}$, then $a(n) < \dfrac{2ln(\sqrt{n})}{1 + \sqrt{n}+ \dfrac{\sqrt{n}^2}{2!} + \dfrac{\sqrt{n}^3}{3!} + \dfrac{\sqrt{n}^4}{4!} + \dfrac{\sqrt{n}^5}{5!}} < \dfrac{2 \sqrt{n}}{\dfrac{\sqrt{n}^5}{5!}}= \dfrac{1}{60n^2} = b(n)$. for n large enough. But the series whose nth term $ b(n) = \dfrac{1}{60n^2}$ converges, and by comparison test, the original series converges.
