Evaluate the limit of the sequence: $\lim_{n_\to\infty}\frac{\sqrt{(n-1)!}}{(1+\sqrt{1})\cdot(1+\sqrt{2})\cdot (1+\sqrt{3})\cdots (1+\sqrt{n})}$ Evaluate the limit of the sequence:
$$\lim_{n\to\infty}\frac{\sqrt{(n-1)!}}{(1+\sqrt{1})\cdot(1+\sqrt{2})\cdot (1+\sqrt{3})\cdots (1+\sqrt{n})}$$

My try:
Stolz-cesaro: The limit of the sequence is $\frac{\infty}{\infty}$
$$\lim_{n\to\infty}\frac{a_n}{b_n}=\lim_{n\to\infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}$$
For our sequence:
$\lim_{n\to\infty}\frac{\sqrt{(n-1)!}}{(1+\sqrt{1})\cdot(1+\sqrt{2})\cdot (1+\sqrt{3})\cdots (1+\sqrt{n})}=\lim_{n\to\infty}\frac{\sqrt{n!}-\sqrt{(n-1)!}}{(1+\sqrt{1})\cdot(1+\sqrt{2})\cdot (1+\sqrt{3})\cdots (1+\sqrt{n})\cdot(1+\sqrt{n+1})-(1+\sqrt{1})\cdot(1+\sqrt{2})\cdot (1+\sqrt{3})\cdots (1+\sqrt{n})}=\lim_{n\to\infty}\frac{\sqrt{(n-1)!}\cdot(\sqrt{n-1})}{\left((1+\sqrt{1})\cdot(1+\sqrt{2})\cdot (1+\sqrt{3})\cdots (1+\sqrt{n})\right)\cdot(\sqrt{n}+1)}$
Which got me nowhere.
 A: $(1+\sqrt{1})\cdot(1+\sqrt{2})\cdot (1+\sqrt{3})\cdots (1+\sqrt{n}) \ge \sqrt{1}\cdot\sqrt{2}\cdots \sqrt{n}= \sqrt{n!}$, hence
$0 \le \frac{\sqrt{(n-1)!}}{(1+\sqrt{1})\cdot(1+\sqrt{2})\cdot (1+\sqrt{3})\cdots (1+\sqrt{n})} \le \frac{1}{\sqrt{n}}.$
Can you proceed ?
A: Consider:
$$
(1+\sqrt{1})\cdot(1+\sqrt{2})\cdot (1+\sqrt{3})\cdots (1+\sqrt{n})
$$
Take the root from each pair of parentheses and multiply them, then:
$$
(1+\sqrt{1})\cdot(1+\sqrt{2})\cdot (1+\sqrt{3})\cdots (1+\sqrt{n}) > \sqrt{n!} \iff \\
\iff \frac{1}{(1+\sqrt{1})\cdot(1+\sqrt{2})\cdot (1+\sqrt{3})\cdots (1+\sqrt{n})} < \frac{1}{\sqrt{n!}}
$$
Going back to original we have that:
$$
\frac{\sqrt{(n-1)!}}{(1+\sqrt{1})\cdot(1+\sqrt{2})\cdot (1+\sqrt{3})\cdots (1+\sqrt{n})} \le \frac{\sqrt{(n-1)!}}{\sqrt{n!}} = \frac{1}{\sqrt n}
$$
But the function is greater than $0$ and hence using squeeze theorem we conclude that:
$$
0 \le \lim_{n\to\infty}x_n \le \lim_{n\to\infty}\frac{1}{\sqrt n} = 0
$$
Hence the limit is $0$.
A: We have that
$$\frac{\sqrt{(n-1)!}}{\prod_{k=1}^n\big(1+\sqrt{k}\big)}=\frac1{\sqrt n}\prod_{k=1}^n\frac{\sqrt{k}}{1+\sqrt{k}}\le \frac1{\sqrt n}$$
A: Note that
$$
\frac{\sqrt{(n-1)!}}{\prod_{k=1}^n\big(1+\sqrt{k}\big)}=\frac{1}{1+\sqrt{n}}
\prod_{k=1}^{n-1}\frac{\sqrt{k}}{1+\sqrt{k}}<\frac{1}{1+\sqrt{n}}\to 0
$$
