# $\lim_{n\to\infty}{\left(\frac12\cdot\frac34\cdot\frac56\cdots\frac{2n-1}{2n}\right)}=0$

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
Yeah I forgot $\cdots$, gonna fix it now. – Lazar Ljubenović Mar 12 '13 at 14:37
Since nobody mentioned the squeeze theorem, I will do it. – J.H. Mar 12 '13 at 15:14
I wouldnt like to open another topic, so I will ask it here: Can someone explain why $A_n < \frac{1}{\sqrt{2n+1}}$? – Giiovanna Oct 13 '14 at 17:24

Yes, since $\displaystyle\frac1{\sqrt{2n+1}}\,\to 0$, so we have $$0<A_n<\frac1{\sqrt{2n+1}}\to 0\,.$$
$$\prod_{m=1}^{n}\frac{2m-1}{2m}= {\frac { \left( \frac{1}{2} \right)^{n+1}{2}^{n+1}\left( n-\frac{1}{2} \right) !}{n! \,\sqrt {\pi }}}.$$
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.