Does this sum diverge? I just can't seem to get my head around this sum: 
$$\sum_{n=1}^\infty \frac{\sqrt{(n!)}}{(2+ \sqrt 1)(2 + \sqrt 2)...(2 + \sqrt n)} $$
The problem I'm having is how to solve the denominator. I tried d'Alembert's ratio test with no success. 
 A: A solution, maybe not the most elegant.
You have
$$
a_n\stackrel{\rm def}{=}\frac{\sqrt{n!}}{\prod_{k=1}^n (2+\sqrt{k})}
= \prod_{k=1}^n \frac{\sqrt{k}}{2+\sqrt{k}}.
$$
Now,
$$
\ln a_n = \sum_{k=1}^n \ln  \frac{\sqrt{k}}{2+\sqrt{k}}= \sum_{k=1}^n \ln  \frac{1}{1+\frac{2}{\sqrt{k}}}.
$$
Since, when $k\to\infty$,  $\ln  \frac{1}{1+\frac{2}{\sqrt{k}}} =\ln  \left(1-\frac{2}{\sqrt{k}}+o\left(\frac{2}{\sqrt{k}}\right)\right)=-\frac{2}{\sqrt{k}}+o\left(\frac{2}{\sqrt{k}}\right)$ and moreover the series $\sum_{k=1}^\infty \frac{2}{\sqrt{k}}$ diverges, we have by theorems of comparison that, when $n\to\infty$,
$$
\ln a_n \operatorname*{\sim}_{n\to\infty} -\sum_{k=1}^n \frac{2}{\sqrt{k}} \operatorname*{\sim}_{n\to\infty} -4\sqrt{n} 
$$
from which $$
a_n = e^{-4\sqrt{n} + o(\sqrt{n})} \tag{$\ast$}
$$
Now, by comparison (with e.g. $b_n\stackrel{\rm def}{=} e^{-n^{0.49}}$, which by $(\ast)$ satisfies $a_n = o(b_n)$ and $\sum_{n=1}^{\infty} b_n < \infty$), we get that the series
$\sum_{n} a_n$ converges.
A: Assuming $n\geq 4$, the general term is given by
$$ a_n=\prod_{k=1}^{n}\frac{\sqrt{k}}{2+\sqrt{k}} = \prod_{k=1}^{n}\left(1+\frac{2}{\sqrt{k}}\right)^{-1}\leq\frac{1}{15}\prod_{k=4}^{n}\left(1+\frac{2}{\sqrt{k}}\right)^{-1}$$
and assuming $n\geq 10$
$$ \prod_{k=4}^{n}\left(1+\frac{2}{\sqrt{k}}\right)\geq 1+\!\!\!\!\!\sum_{\substack{A\subseteq\{\lfloor n/2\rfloor-1,\ldots,n\}\\|A|=4}}\prod_{a\in A}\frac{2}{\sqrt{a}}\geq\binom{n/2}{4}\frac{16}{n^2}\geq\frac{1}{24}\left(\frac{n}{2}-3\right)^2 $$
so the given series is convergent by comparison with $\sum_{n\geq 1}\frac{1}{n^2}$, that is convergent.
