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Prove that $$ \lim_{n\to\infty}{\left(\frac12\cdot\frac34\cdot\frac56\cdot\ldots\cdot\frac{2n-1}{2n}\right)}=0. $$

Transforming it to factorial obviously doesn't help at all, so I've noted $A_n$ as above product and noticed $1/2<2/3$, $3/4<4/5$, ..., $(2n-1)/(2n)<(2n)/(2n+1)$, so $A_n<1/\sqrt{2n+1}$.

Now it's kind of obvious that $A_n$ approached $0$ as $n$ approaches infinity, but I'm not sure about formality of this proof. Is it safe to conclude that $A_n\to0^+ \text{when } n\to\infty$ from $A_n>0 \land A_n<1/\sqrt{2n+1}=0^+ \text{when } n\to\infty$?

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The way you've written it now, it actually tends to 5/16. –  Joe Z. Mar 12 '13 at 14:37
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Yeah I forgot $\cdots$, gonna fix it now. –  Lazar Ljubenović Mar 12 '13 at 14:37
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Since nobody mentioned the squeeze theorem, I will do it. –  J.H. Mar 12 '13 at 15:14
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2 Answers

up vote 3 down vote accepted

Yes, since $\displaystyle\frac1{\sqrt{2n+1}}\,\to 0$, so we have $$0<A_n<\frac1{\sqrt{2n+1}}\to 0\,.$$

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If you put some effort, you can have the following identity

$$ \prod_{m=1}^{n}\frac{2m-1}{2m}= {\frac { \left( \frac{1}{2} \right)^{n+1}{2}^{n+1}\left( n-\frac{1}{2} \right) !}{n! \,\sqrt {\pi }}}.$$

Using Stirling approximation for $ n! \sim \sqrt{2 \pi n} \left(\frac{n}{e}\right)^n $ and taking limit as $n$ goes to infinity, the desired result follows.

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