Another evaluating limit question: $\lim\frac{1\cdot3\cdot5\cdot\ldots\cdot(2n-1)}{2\cdot4\cdot6\cdot\ldots\cdot2n}$

How do I begin to evaluate this limit: $$\lim_{n\to \infty}\ \frac{1\cdot3\cdot5\cdot\ldots\cdot(2n-1)}{2\cdot4\cdot6\cdot\ldots\cdot2n}\;?$$

Thanks a lot.

Solution 1:

A simple, but famous trick works here: Observe that $(n-1)(n+1) = n^2 - 1 \leq n^2$. Thus we have $$ \begin{align*} & \left[ \frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{2 \cdot 4 \cdot 6 \cdots (2n)} \right]^2 \\ &= \frac{1 \cdot 1}{2 \cdot 2}\cdot\frac{3 \cdot 3}{4 \cdot 4}\cdot\frac{5 \cdot 5}{6 \cdot 6}\cdots\frac{(2n-3) \cdot (2n-3)}{(2n-2) \cdot (2n-2)}\cdot\frac{(2n-1) \cdot (2n-1)}{(2n) \cdot (2n)} \\ &= \frac{1 \cdot 3}{2 \cdot 2}\cdot\frac{3 \cdot 5}{4 \cdot 4}\cdot\frac{5 \cdot 7}{6 \cdot 6}\cdots\frac{(2n-3) \cdot (2n-1)}{(2n-2) \cdot (2n-2)}\cdot\left(\frac{2n-1}{(2n)^2}\right) \\ & \leq \frac{2n-1}{(2n)^2} \\ & \leq \frac{1}{2n}. \end{align*} $$ Thus we have $$\frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{2 \cdot 4 \cdot 6 \cdots (2n)} \leq \frac{1}{\sqrt{2n}} $$ and the limit is zero. In fact, the sharp estimate $$ \frac{1}{\sqrt{(\pi + o(1)) n}} \leq \frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{2 \cdot 4 \cdot 6 \cdots (2n)} \leq \frac{1}{\sqrt{\pi n}} $$ holds, so the estimation above is not so far from the truth.

Solution 2:

This is almost a quite famous limit.

Wallis showed that

$$\lim_{n \to \infty} \left(\frac{(2n)!!}{(2n-1)!!}\right)^2 \frac 1 n = \pi$$

or that

$$ \frac{(2n)!!}{(2n-1)!!} \sim \sqrt{\pi n}$$

Your limit is

$$\lim_{n \to \infty} \frac{(2n-1)!!}{(2n)!!}$$

which by the above asymtotical behaviour is

$$\lim_{n \to \infty} \frac{(2n-1)!!}{(2n)!!} = \lim_{n \to \infty} \frac{1}{\sqrt{\pi n}}=0$$

Note you can write you expression as

$$\frac{{\left( {2n - 1} \right)!!}}{{\left( {2n} \right)!!}} = \frac{{\left( {2n - 1} \right)!!}}{{\left( {2n} \right)!!}}\frac{{\left( {2n} \right)!!}}{{\left( {2n} \right)!!}} = \frac{{\left( {2n} \right)!}}{{{4^n}n{!^2}}}$$

so using Stirling's approximation, one has

$$\eqalign{ & \left( {2n} \right)! \sim {\left( {\frac{{2n}}{e}} \right)^{2n}}2\sqrt {n\pi } \cr & n{!^2} \sim {\left( {\frac{n}{e}} \right)^{2n}}2n\pi \cr} $$

from where

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

as it was previously stated.

In general,

$${\left[ {\frac{{\left( {2n} \right)!!}}{{\left( {2n - 1} \right)!!}}} \right]^2}\left( {1 - \frac{1}{{2n + 1}}} \right)\frac{1}{n} < \pi < {\left[ {\frac{{\left( {2n} \right)!!}}{{\left( {2n - 1} \right)!!}}} \right]^2}\frac{1}{n}$$

$$\left[ {\frac{{\left( {2n} \right)!!}}{{\left( {2n - 1} \right)!!}}} \right]\sqrt {1 - \frac{1}{{2n + 1}}} < \sqrt {n\pi } < \left[ {\frac{{\left( {2n} \right)!!}}{{\left( {2n - 1} \right)!!}}} \right]$$

Solution 3:

Well known equality : $$\frac{1}{2\sqrt{n}} \leqslant \frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} \cdots \frac{2n-1}{2n}\leqslant \frac{1}{\sqrt{2n}}$$ Proof of right side. Having $$\frac{1}{2}<\frac{2}{3},\frac{3}{4}<\frac{4}{5},\frac{5}{6}<\frac{6}{7},\cdots,\frac{2n-3}{2n-2}<\frac{2n-2}{2n-1},\frac{2n-1}{2n}<1$$ and multiplying we obtain $$\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} \cdots \frac{2n-1}{2n}<\frac{2}{3}\cdot \frac{4}{5}\cdot \frac{6}{7} \cdots \frac{2n-2}{2n-1}$$ Now multiply both sides on $ \frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} \cdots \frac{2n-1}{2n}$ gives $$ \left(\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} \cdots \frac{2n-1}{2n}\right)^2<\frac{1}{2n}$$ and we obtain $$\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} \cdots \frac{2n-1}{2n} <\frac{1}{\sqrt{2n}}$$ Analogical way works for left side.