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. 
 A: 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.
A: 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.)
A: 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 $0$.
A: Note that $\log(x) = 2\log\sqrt{x}$. Then, we have
$$ 2\sum_{n=1}^\infty \log(\sqrt{n})\mathrm{e}^{-\sqrt{n}}. $$
This sort of looks like it could be integrable (!), so let us recall that the sum "does the same thing" as the corresponding integral. Letting $u=\sqrt{x}$, we have
$$ 4\int_1^\infty u\log(u) \mathrm{e}^{-u}\,du. $$
At this point, there are any number of things we could do. We could try taking the integral exactly, but that will probably be nasty. It was remarked earlier that $\log{x}<x$ in this domain; this seems like the smart trick to use. Our integral will then be bounded above by
$$ 4\int_1^\infty u^2\mathrm{e}^{-u}\,du < 4\int_0^\infty u^2 \mathrm{e}^{-u}\,du = 4\Gamma(3) = 4\cdot2! = 8. $$
Any excuse to use the definition of the gamma function, but more important, we demonstrated that the integral is bounded above by a fixed constant. This, combined with the observation that every individual term is positive (so the sequence of partial sums is monotonic) gives us the convergence result we want -- namely that by the integral test, then some craft inequalities, we see that the series must be convergent.
