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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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closed as unclear what you're asking by ᴡᴏʀᴅs, MathOverview, Zachary Selk, S.Panja-1729, Ivo Terek Oct 11 '15 at 2:34

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question.If this question can be reworded to fit the rules in the help center, please edit the question.

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
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
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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