Showing that $\lim_{n\to\infty}\frac{1\cdot 3\cdot 5\cdots (2n-1)}{(2n)^n} = 0$

Question

Ok, so I want to show that

$$\lim_{n\to\infty}\frac{1\cdot 3\cdot 5\cdots (2n-1)}{(2n)^n} = 0.$$

Here is what I have tried so far:

\begin{align} \notag \lim_{n\to\infty}\frac{1\cdot 3\cdot 5\cdot \cdots \cdot (2n-1)}{(2n)^n} &= \lim_{n\to\infty}\frac{\prod_{j=1}^{n}(2j-1)}{(2n)^n}\\ \notag &= \lim_{n\to\infty}\frac{\prod_{j=1}^{n}(2j-1)}{(2n)^n}\cdot\frac{\prod_{j=1}^{n}(2j)}{\prod_{j=1}^{n}(2j)}\\ \notag &= \lim_{n\to\infty}\frac{\prod_{j=1}^{2n}j}{(2n)^n\prod_{j=1}^{n}(2j)}\\ \notag &= \lim_{n\to\infty}\frac{(2n)!}{(2n)^n\prod_{j=1}^{n}(2j)}\\ \notag &= \lim_{n\to\infty}\frac{(2n)!}{(2n)^n\cdot 2^n\cdot n!}\\ \notag &= \lim_{n\to\infty}\frac{(n+1)\cdot\cdots\cdot (2n)}{(4n)^{n}} \end{align}

From here you can tell that it vanishes, but is this enough to show that the limit is 0? Is there any way to break it down further? Additionally, are there any handy product identities to know? Basically anything analogous to identities like the following:

$$\sum_{i=1}^n = \frac{n(n+1)}{2}$$

Thanks

Answer 1

HINT:

$$\frac{1\cdot 3\cdot 5\cdots (2n-1)}{(2n)^n}=\prod_{1\le r\le n}\frac{2r-1}{2n}<\frac1{2n} $$

as each term $\le1$ and $2n>2r-1$ for finite $r$

Answer 2

What about this? $$ \lim_{n \to \infty} \frac{1 \cdot 3 \cdot \dots \cdot 2n-1}{(2n)(2n) \cdots (2n)} \le \lim_{n \to \infty} \frac 1{2n} \cdot 1 \cdot \dots \cdot 1 = 0 $$

Answer 3

Another approach:

$$a_n:=\frac{1\cdot 3\cdot\ldots\cdot(2n-1)}{2^nn^n}=\frac{(2n)!}{2^{2n}n!n^n}$$

so

$$\frac{a_{n+1}}{a_n}=\frac{(2n+2)!}{2^{2n+2}(n+1)!(n+1)^{n+1}}\frac{2^nn!n^n}{(2n)!}=\frac12\frac{2n+1}{n+1}\frac1{\left(1+\frac1n\right)^n}\xrightarrow[n\to\infty]{}\frac1e<1\implies$$

$$\implies\sum_{n=1}a_n\;\;\text{converges}\;\implies a_n\xrightarrow[n\to\infty]{}0$$

Answer 4

One more way: rewrite numerator as $(2n-1)!!=\frac{(2n)!}{2^{n}n!}$ and denominator using Stirling's formula as $(2n)^n=(2n)^{\frac{2n}{2}}=e^{2n} (2n)! 2\sqrt{\pi n}(1+o(1))$. Hence your expression becomes $$ S=\frac{(2n)!}{2^n n! e^{2n} (2n)! 2\sqrt{\pi n}(1+o(1))}=\frac{1}{2^n n! e^{2n} (2n)! 2\sqrt{\pi n}(1+o(1))} $$ which happily tends to $0$.

Answer 5

My hint:

$$\frac{(2n-1)!!}{(2n)^n}=\frac{(2n)!}{n!(4n)^n}\sim \frac{(\frac{2n}{e})^{2n}\sqrt{4\pi n}}{(\frac{n}{e})^{n}\sqrt{2\pi n}(4n)^n}=\frac{\sqrt{2}(2n)^n}{(2ne)^n}=\sqrt{2}\: \left(\frac{1}{e}\right)^n\to 0$$

Hence: $$\lim_{n\to \infty}\frac{(2n-1)!!}{(2n)^n}=0$$

I use Stirling's approximation