<s>[INST] Let $a_{n}=(1-\frac{1}{\sqrt{2}})\ldots(1-\frac{1}{\sqrt{n+1}}), n \geq 1 $. Then $\lim_\limits{n \rightarrow \infty} a_{n} $ Options are  (A) equals 1 (B) does not exist (C) equals $\frac{1}{\sqrt{\pi}}$ (D) 0.
I have multiplied each numerator with $\sqrt{n+1}+1$, $\forall n = 1$ to $\infty$ and then got $a_{n}=\frac{n!}{(2+\sqrt{2})(3+\sqrt{3})\ldots((n+1)+\sqrt{n+1})}$. So this must converge to 0 right since the fraction comes as $\frac{n!}{(n+1)!+\mathrm{something}}$ ?
 [/INST] We can write
$$a_n = \frac{\sqrt{2}-1}{\sqrt{2}}\frac{\sqrt{3}-1}{\sqrt{3}}\cdots \frac{\sqrt{n+1}-1}{\sqrt{n+1}} = \frac{1}{\sqrt{2}(\sqrt{2}+1)}\frac{2}{\sqrt{3}(\sqrt{3}+1)}\cdots \frac{n}{\sqrt{n+1}(\sqrt{n+1}+1)}$$
For every positive integer $k$, $\sqrt{k}(\sqrt{k}+1) > \sqrt{k}\cdot\sqrt{k} = k$; therefore
$$a_n < \frac{1}{2}\frac{2}{3}\cdots \frac{n}{n+1}=\frac{1}{n+1}$$
Since $\lim \frac{1}{n+1} = 0$ and $a_n$ is positive, by the squeeze theorem we must have $\lim a_n = 0$.
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