calculus - Evaluate the limit of the sequence: $lim_{n_toinfty}frac{sqrt{(n-1)!}}{(1+sqrt{1})cdot(1+sqrt{2})cdot (1+sqrt{3})cdots (1+sqrt{n})}$

Evaluate the limit of the sequence:

$$\lim_{n\to\infty}\frac{\sqrt{(n-1)!}}{(1+\sqrt{1})\cdot(1+\sqrt{2})\cdot (1+\sqrt{3})\cdots (1+\sqrt{n})}$$

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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.

Answer

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 $0$.