What is the limit of $\frac{\prod\mathrm{Odd}}{\prod\mathrm{Even}}?$ Does $\pi$ show up here? 
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?
 A: To paraphrase Wikipedia: if $a_n$ is a sequence of complex numbers such that $\sum_{n\geq 1} |a_n|^2 < \infty$, then non-zero convergence of $\prod_{n\geq 1}(1+a_n)$ is equivalent to convergence of $\sum_{n\geq 1} a_n$.
Letting $a_n = -1/2n$, we can see that, because the harmonic series diverges to infinity, your product converges to zero.
A: If $0\leq a_n<1$ for all $n,$ then $$\prod_{n=1}^{\infty}(1-a_n)=0\iff \sum_{n=1}^{\infty}a_n=\infty.$$ Let $a_n=1/2 n.$ Then $\sum_{n=1}^{\infty}a_n=\infty.$ .
So $\quad \prod_{n=1}^{\infty}\frac {2 n-1}{2 n}=\prod_{n=1}^{\infty}(1-a_n)=0.$
A: Your product is mostly understood here as :
$$P^*= \lim_{n \to \infty} \prod_{k=1}^n (2k-1)/2k = \lim_{n \to \infty} \prod_{k=1}^n (1- \frac1{2k})  $$

My intuition how to interpret your product was rather $$P= \lim_{n \to \infty} \prod_{k=1}^n k^{-(-1)^k}$$ and for this interpretation a regularization on the formal logarithm can be applied, namely Cearosummation  $\mathfrak C$ for the logarithms of the product-terms and this gives indeed a relation to $\pi$:
 $$\log(P) = L \underset{\mathfrak C}=\sum_{k=1}^\infty (-1)^{k-1} \cdot \log(k) \approx -0.225791352645 $$
The exponential is then 
 $$ P = \exp(L) = \sqrt{\frac2\pi}$$

Using Pari/GP:
L = -sumalt(k=1 , (-1)^k * log(k) )   \\ "sumalt()" provides sort of Cesarosum
print([ P=exp(L) , C = 1/sqrt(Pi/2)  , err = P - C ])
 \\ result:
 [0.797884560803, 0.797884560803, -1.005331151 E-200]

Remark: if that summation-method is allowed then the result is also expressible as the (regularized) derivative of the Dirichlet's eta-function at zero: (I've implemented the eta-function in Pari/GP using the Euler's conversion from the zeta:      
$$ L = - \eta(0)'  \\
  P = e^{- \eta(0)' }$$

While you evaluate your product as partial products $ P_{1,n} = \frac12 \cdot \frac34 \cdots \frac {2n-1}{2n} $ one can do as well with $P_{2,n} = 1 \cdot \frac 32 \cdot \frac54 \cdots \frac{2n+1}{2n}  $. While $\lim_{n \to \infty }P_{1,n}$ converges to zero diverges $P_{2,n}$ (to infinity).
But interestingly, if we take the partial evaluation up to the same $n$ , then we do $P = \lim_{n \to \infty} \sqrt{P_{1,n} \cdot P_{2,n}}$ with $P$ having the same value as given in my example using Cesaro-/Euler-summation.      

We encounter here the same problem as when we sum the Grandi-series, with parentheses around pairs of terms in the two possible ways and get two different values: but when we take the mean of that two values we get the Cesaro-sum.
A: 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 $1/ \sqrt \pi$ 
A: Here you need to find $\lim A_n$, where $$A_n=
\frac{1\times3\times5\times\cdots\times(2n-1)}{2\times4\times6\times8\times\cdots\times (2n)}.$$
Now note that, for $n\gt1$, $2n-1\gt n$. So, in the numerator, you get $$\frac{1\times 2\times 3\dots\times n}{2\times4\times 6\times\dots\times 2n}\lt\frac{1\times3\times5\times\cdots\times(2n-1)}{2\times4\times6\times8\times\cdots\times (2n)}\\\implies\frac1{2^n}\lt \frac{1\times3\times5\times\cdots\times(2n-1)}{2\times4\times6\times8\times\cdots\times (2n)}$$
Now, you use the identity $$\frac n{n+1}\lt\frac {n+1}{n+2}$$ and define a sequence $$B_n=\frac 23\times \frac 45\times\dots \times \frac {2n}{2n+1} .$$
So, now you have $$A_n\lt B_n\\\implies (A_n)^2\lt A_nB_n=\frac 1{2n+1}\\\implies A_n\lt \frac 1{\sqrt{2n+1}}.$$
So, you have $$\frac 1{2^n}\lt A_n\lt \frac 1{\sqrt{2n+1}}\\\implies\lim \frac 1{2^n}\lt\lim A_n\lt\lim \frac 1{\sqrt{2n+1}}.$$
So, by sandwitch theorem, $$\lim A_n=0.$$
A: As has been pointed out, the limit is zero.
However you mention that you remember it's something involving $\pi$.  What you're probably thinking of is something like the result
$$\lim_{n \to \infty} \sqrt{n} {1 \cdot 3 \cdots (2n-1) \over 2 \cdot 4 \cdots  2n} = \frac 1{\sqrt{\pi}}.$$
Notice the additional factor of $\sqrt{n}$ in front.  This means that your initial product, with $n$ factors on the top and $n$ factors on the bottom, is near $1/{\sqrt{\pi n}}$ when $n$ is larger.  For example if $n = 100$ the product is about 0.05634 and $1/\sqrt{100\pi} \approx 0.05642$.
The limit quoted above can be proved using Stirling's formula,
$$\lim_{n \to \infty} {n! \over \sqrt{2\pi n} (n/e)^n} = 1, $$
if you rewrite the product as $(2n)!/(2^n n!)^2$.  
A: 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 +$\infty$. 
