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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.

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See convergence criteria for infinite products. Also, note that an infinite product is said to diverge to $0$ – Lucian Apr 16 '14 at 18:28
@Lucian I've never heard of anything diverging on $0$. I'm sure you meant diverging to $\infty$ (or converge to $0$) – Cruncher Apr 16 '14 at 19:14
@Cruncher: Check the link. It also contains an explanation. – Lucian Apr 16 '14 at 19:19

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.

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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 $$\lim_{n\to\infty} \frac{\sqrt{n!}}{\prod\limits_{k=1}^n (1+\sqrt{k})} = 0$$

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How about this? – Anastasiya-Romanova 秀 Apr 19 '14 at 20:56

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.

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@GregVoigt Your proof is not correct. The statement does not follow from the fact that $\lim_{n\to\infty}\frac1{\sqrt{n}}=0$ – Etienne Apr 16 '14 at 18:41
@GregVoigt I am not the downvoter... – Etienne Apr 16 '14 at 18:48

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: $$ \prod_{k=1}^n \frac{\sqrt{k}}{\sqrt{k}+1} < \frac{1}{\sqrt{n}} \to 0$$

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