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$\displaystyle \lim_{n \rightarrow \infty}\frac{1 \cdot 3 \cdot 5 \dots \cdot (2n-1)}{2 \cdot 4 \cdot 6 \cdot \dots \cdot (2n)} $

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What have you tried? –  Stefan Nov 29 '12 at 16:30
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It is not all that easy. The limit is $0$, but the decay rate is not fast. A quick and perhaps wrong calculation gives size about $\frac{C}{\sqrt{n}}$ for some constant $C$. In particular, for the alternating series I think you asked recently about, we have convergence but not absolute convergence. –  André Nicolas Nov 29 '12 at 19:25

4 Answers 4

up vote 2 down vote accepted

$$\dfrac12 \cdot \dfrac34 \cdot \dfrac78 \cdots \dfrac{2n-1}{2n} = \left(1 - \dfrac12\right)\left(1 - \dfrac14\right)\left(1 - \dfrac16\right)\left(1 - \dfrac18\right)\cdots\left(1 - \dfrac1{2n}\right)$$ Since $$\dfrac12 + \dfrac14 + \dfrac16 + \cdots + \dfrac1{2n} + \cdots$$ diverges, the infinite product goes to $0$.

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You are using that $\prod (1-a_n) = 0$ $\Leftrightarrow$ $\sum a_n=\infty$, right? –  Martin Sleziak Nov 30 '12 at 7:33
    
Yes. If $a_n < 1$, $\sum a_n = \infty \iff \prod (1-a_n) = 0$. –  user17762 Nov 30 '12 at 7:34
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Since I am not able to find a better reference quickly, I'll mention at least the first comment in this blog post. –  Martin Sleziak Nov 30 '12 at 7:35

There are all sorts of arguments. I just want to say that this infinite product originates from a proof of the Wallis product formula for $\pi$, in which one can show, using integration by parts, that $$\int_0^{\pi/2} \sin^{2n}(x)dx = \frac{\pi}{2}\frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n)}.$$

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A hint:

Write $1\cdot3\cdot 5\cdot\ldots\cdot(2n-1)$ and $2\cdot 4\cdot 6\cdot\ldots\cdot (2n)$ in terms of factorials and powers of $2$. Then use Stirling's formula

$$m!=\left({m\over e}\right)^m\ \sqrt{2\pi m}\ \bigl(1+o(1)\bigr)\qquad(m\to\infty)$$

to estimate the various factorials appearing in your expression.

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Note that $$\frac{{1.3{\cdots}(2n - 1)}}{{2.4{\cdots}(2n)}} = \frac{{\frac{{(2n - 1)!}}{{2.4{\cdots}(2n - 2)}}}}{{{2^n}n!}} = \frac{{\frac{{(2n - 1)!}}{{{2^{n - 1}}(n - 1)!}}}}{{{2^n}n!}} = \frac{{(2n - 1)!}}{{{2^{n - 1}}(n - 1)!}} \cdot \frac{1}{{{2^n}n!}} \to 0$$ using Stirling's approximation because $n$ is "large" here.

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