What is the limit of $\frac{\prod\mathrm{Odd}}{\prod\mathrm{Even}}?$ What is this limit
  $$
  \frac{1\times3\times5\times\cdots}{2\times4\times6\times8\times\cdots} = \lim_{n \rightarrow \infty}\prod_{i=1}^{n}\frac{(2i-1)}{2i}
$$


I remember that it was something involving $\pi$.

How can I compute it?



In addition; how can I compute it's sum series limit?
If you take the numerator product up to $2n-1$ and the denominator up to $2n,$ you have exactly 
$$ \frac{(2n)!}{4^n (n!)^2}  $$

By Stirling's approximation this is asymptotic to
$$ \frac{1}{\sqrt {\pi n}} $$
and goes to $0$ slowly. If we were to add one more term to the numerator, that is $2n+1,$ the expression would go to $\infty$ slowly, as a constant times $\sqrt n.$

The only simple way to get a limit is to take the numerator up to $2n-1$ as before, but also multiply the numerator by a single factor of $\sqrt n.$ Then the limit would be $1/ \sqrt \pi$
Here's a formula which I found embedded in an old C program. I don't know where this comes from, but it converges to Pi very quickly, about 16 correct digits in just 22 iterations:

$\pi = \sum_{i=0}^{\infty}{
\frac{6(\prod{2j-1})}
{(\prod{2j})(2i+1)(2^{2i+1})}}$

(Each product is for j going from 1 to i. When i is 0, the products are empty, equivalent to 1/1. When i is 1, the products are 1/2. When i is 2, the products are 3/4. Etc.)

The limit of the quotient of the products is 0, but the limit of the sum as I give it above is exactly equal to $\pi$.

As for the limit of the sum of the quotients of the products, it diverges to +$\infty$.