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?

Solution 1:

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 $1/ \sqrt \pi$

Solution 2:

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, $$\lim A_n=0.$$