Limit $\lim_{n \rightarrow \infty}\frac{1 \cdot 3 \cdot 5 \dots \cdot (2n-1)}{2 \cdot 4 \cdot 6 \cdot \dots \cdot (2n)} $ $\displaystyle  \lim_{n \rightarrow \infty}\frac{1 \cdot 3 \cdot 5 \dots \cdot (2n-1)}{2 \cdot 4 \cdot 6 \cdot \dots \cdot (2n)} $ 
 A: 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.
A: $$\dfrac12 \cdot \dfrac34 \cdot \dfrac78 \cdots \dfrac{2n-1}{2n} = \left(1 - \dfrac12\right)\left(1 - \dfrac14\right)\left(1 - \dfrac16\right)\left(1 - \dfrac18\right)\cdots\left(1 - \dfrac1{2n}\right)$$
Since $$\dfrac12 + \dfrac14 + \dfrac16 + \cdots + \dfrac1{2n} + \cdots$$ diverges, the infinite product goes to $0$.
A: 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.
A: 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
$$\int_0^{\pi/2} \sin^{2n}(x)dx = \frac{\pi}{2}\frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n)}.$$
