Limit of probability of getting n 1's in 2n random bits

Question

Suppose we generate $2n$ random bits. Let $p_n$ be the probability of getting exactly $n$ 1's. I want to find $\lim_{n\to \infty} p_n$. I have $$p_n = \frac{\binom{2n}{n}}{2^{2n}} = \frac{(2n)!}{(n!)^2 2^{2n}}$$ and $$(2n)! = \prod_{i=1}^n (2i) \prod_{i=1}^n (2i-1) = \left[ \prod_{i=1}^n (2i)\right]^2 \frac{\prod_{i=1}^n (2i-1)}{\prod_{i=1}^n (2i)} = (2^n n!)^2 \prod_{i=1}^n\frac{ 2i-1}{2i}$$ So $$p_n = \prod_{i=1}^n\frac{ 2i-1}{2i}$$ So $$\lim_{n\to \infty}p_n = \prod_{i=1}^\infty\frac{ 2i-1}{2i}$$ What is the value of this infinite product?

Answer 1

By Stirling's approximation:

$$n!=\sqrt{2\pi n}\left(\frac{n}{e}\right)^n\left(1+O\left(\frac1n\right)\right)$$

Note that as $n$ approaches $\infty$, $O\left(\frac1n\right)$ drops to $0$.

Using this approximation:

$$\lim_{n\to\infty}\frac{(2n)!}{(n!)^2\cdot2^{2n}}$$

$$=\lim_{n\to\infty}\frac{\sqrt{4\pi n}\left(\frac{2n}{e}\right)^{2n}\left(1+O\left(\frac1n\right)\right)}{\left(\sqrt{2\pi n}\left(\frac{n}{e}\right)^{n}\left(1+O\left(\frac1n\right)\right)\right)^2\cdot2^{2n}}$$

$$=\lim_{n\to\infty}\frac{\sqrt{4\pi n}\left(\frac{2n}{e}\right)^{2n}}{\left(\sqrt{2\pi n}\left(\frac{n}{e}\right)^{n}\right)^2\cdot2^{2n}}$$

$$=\lim_{n\to\infty}\frac{2\sqrt{\pi n}\frac{(2n)^{2n}}{e^{2n}}}{2\pi n\frac{n^{2n}}{e^{2n}}\cdot2^{2n}}$$

$$=\lim_{n\to\infty}\frac{2^{2n}\cdot n^{2n}}{\sqrt{\pi n}\cdot n^{2n}\cdot2^{2n}}$$

$$=\lim_{n\to\infty}\frac1{\sqrt{\pi n}}$$

$$=0$$

$\therefore$ As $n$ approaches infinity, the chance that $2n$ bits contains $n$ 1s approaches $0$.

Answer 2

Finishing it off:

$\prod_{i=1}^\infty \frac{2i-1}{2i} = \prod_{i=1}^\infty (1 - \frac{1}{2i}) \leq \prod_{i=1}^\infty \exp(-\frac{1}{2i}) = \exp(-\frac{1}{2} \sum_{i=1}^\infty \frac{1}{i}) = 0$

since the harmonic series diverges.