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.
Best 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 $0$.