How to show that this limit is tend to zero? How to show that this limit is tend to zero?
$$\lim_{n\to\infty}\frac{\sqrt{n!}}{(1+\sqrt{1})(1+\sqrt{2})\cdots(1+\sqrt{n})}=0$$
Thank you.
 A: Notice 
$$k = (k-1)+1 \le (\sqrt{k-1}+1)^2\quad\implies\quad\sqrt{k} \le \sqrt{k-1}+1,$$
we have
$$0 \le \frac{\sqrt{n!}}{\prod\limits_{k=1}^n (1+\sqrt{k})} = \prod_{k=1}^n \frac{\sqrt{k}}{\sqrt{k}+1} \le \prod_{k=1}^n \frac{\sqrt{k-1}+1}{\sqrt{k}+1} = \frac{1}{\sqrt{n}+1}$$
Since the RHS goes to $0$ as $n \to \infty$, we find
$$\lim_{n\to\infty} \frac{\sqrt{n!}}{\prod\limits_{k=1}^n (1+\sqrt{k})} = 0$$
A: You have 
$$\frac{\sqrt{n!}}{(1+\sqrt 1)(1+\sqrt 2)\cdots (1+\sqrt{n})}=\frac1{\left(1+\frac1{\sqrt1}\right)\left(1+\frac1{\sqrt2}\right)\cdots \left(1+\frac1{\sqrt {n}}\right)}\cdot $$
So you need to show that 
$$\lim_{n\to\infty} \prod_{k=1}^n \left(1+\frac1{\sqrt{k}}\right)=\infty\, ; $$
in other words, that the infinite product $\prod \left(1+\frac1{\sqrt{k}}\right)$ is divergent to $\infty$. 
This is true because the series $\sum\frac1{\sqrt{k}}$ is divergent.
A: You can rewrite your expression as
$\lim_{n\rightarrow\infty}\Pi_{i=1}^{n}\frac{\sqrt{i}}{1+\sqrt{i}}=
\\
\lim_{n\rightarrow\infty}\Pi_{i=1}^{n}\frac{1}{\frac{1}{\sqrt{i}}+1}
$
Now, we know that $\lim_{n\rightarrow\infty}\frac{1}{\sqrt{n}}=0$. You can go from there.
A: Other ways to increase:
1)$$0 < \frac{\sqrt{n!}}{\prod\limits_{k=1}^n (1+\sqrt{k})} = \prod_{k=1}^n \frac{\sqrt{k}}{\sqrt{k}+1} < \prod_{k=1}^n \sqrt{\frac{k}{k+1}} = \frac{1}{\sqrt{n+1}} \to 0$$
2) Apply mathematical induction can be demonstrated:
$$ \prod_{k=1}^n \frac{\sqrt{k}}{\sqrt{k}+1} < \frac{1}{\sqrt{n}} \to 0$$
