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 $\frac{n!}{(n+1)!+\mathrm{something}}$ ?
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Rudin Real and Complex Analysis 15.5: "Suppose $0\le u_n<1$. Then $\prod_{n=1}^\infty (1-u_n)>0$ if and only if $\sum_{n=1}^\infty u_n<\infty$." – ForgotALot Apr 19 '16 at 03:11
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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 $\lim a_n = 0$.
kobe
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HINT:
$$\prod_{k=1}^n\left(1-\frac{1}{\sqrt{k+1}}\right)=e^{\sum_{k=1}^n \log\left(1-\frac{1}{\sqrt{k+1}}\right)}$$
and $\log\left(1-\frac{1}{\sqrt{k+1}}\right)=-\frac{1}{\sqrt{k+1}}+O\left(\frac{1}{(k+1)^{3/2}}\right)$. Now, apply the comparison test to see that the coveted limit must be $0$.
Mark Viola
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