Verification of limit of $\frac{n!}{\prod_{i=1}^n(2i-1)}$

Question

Does the limit $$\lim_{n\to\infty}\frac{n!}{\prod_{i=1}^n(2i-1)}$$ equal zero?

I definitely think no, because, the sequence is divergent as the the sequence is equal to $\prod_{i=1}^n2i=2^nn!$, which is divergent right? On the other hand, does Cauchy theorem on limits work here?

Answer 1

Since $$ \prod_{i=2}^{n}(2i-2)\leq\prod_{i=1}^{n}(2i-1)\leq \prod_{i=1}^{n}2i, $$ we have $$\label{1}\tag{1} \frac{n!}{\prod_{i=1}^{n}(2i-1)}\geq \frac{n!}{\prod_{i=1}^{n}2i}=\frac{n!}{2^{n}n!}=\frac{1}{2^{n}} $$ and $$\label{2}\tag{2} \frac{n!}{\prod_{i=1}^{n}(2i-1)}\leq \frac{n!}{\prod_{i=2}^{n}(2i-2)}=\frac{n!}{\prod_{i=2}^{n}2(i-1)}=\frac{n!}{2^{n-1}(n-1)!}=\frac{n}{2^{n-1}}. $$ Since the right hand side of \eqref{1} and \eqref{2} tends to zero, the sandwich theorem implies that $$ \lim_{n\rightarrow\infty}\frac{n!}{\prod_{i=1}^{n}(2i-1)}=0. $$

Answer 2

By the Stirling approximation, your product $$\frac{n!^2 2^n}{(2n)!}\approx\frac{2\pi n^{2n+1}(2e^{-2})^n}{\sqrt{4\pi n}2^{2n}n^{2n}e^{-2n}}=\frac{\sqrt{\pi n}}{2^n}\stackrel{n\to\infty}{\to}0.$$Or if you only want to know that it vanishes rather than how quickly, note that $$a_n:=\frac{n!}{\prod_{i=1}^n(2i-1)}\implies\frac{a_{n+1}}{a_n}=\frac{n+1}{2n+1}\stackrel{n\to\infty}{\to}\frac12,$$and in particular this ratio $<\frac{2}{3}$ for $n\ge 3$, so there's an exponentially decaying upper bound on $a_n$.

Answer 3

Hint:

Just consider for $a_n \frac{n!}{\prod_{i=1}^n(2i-1)}$ the quotient

$$\frac{a_{n+1}}{a_n} = \frac{(n+1)!}{\prod_{i=1}^{n+1}(2i-1)}\cdot \frac{\prod_{i=1}^n(2i-1)}{n!} = \frac{n+1}{2n+1}\stackrel{n \to \infty}{\longrightarrow}\frac{1}{2}$$

Now, you have for $N$ large enough vor all $n \geq N$: $$a_n \leq C\left(\frac{3}{4}\right)^{n-N}$$

Answer 4

$$a_n=\frac{n!}{\prod_{i=1}^n(2i-1)}=\frac{\sqrt{\pi }\, 2^{-n} n!}{\Gamma \left(n+\frac{1}{2}\right)}$$

Taking the logarithms, using Stirling approximation and continuing with Taylor series $$\log(a_n)=-n \log (2)+\frac{1}{2} \left(\log (\pi )+\log \left({n}\right)\right)+\frac{1}{8 n}+O\left(\frac{1}{n^3}\right)$$ making $$a_n \sim 2^{-n}\sqrt{n \pi}$$

Try for $n=10$; the exact value is $\frac{256}{46189}\approx 0.00554245$ while the above approximation gives $\frac{1}{512} \sqrt{\frac{5 \pi }{2}}\approx 0.00547362$.

Answer 5

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{\lim_{n \to \infty}{n! \over \prod_{i = 1}^{n}\pars{2i - 1}}} = \lim_{n \to \infty}{n! \over 2^{n}\prod_{i = 1}^{n}\pars{i - 1/2}} = \lim_{n \to \infty}{n! \over 2^{n}\pars{1/2}^{\large\overline{n}}} \\[5mm] = &\ \lim_{n \to \infty}{n! \over 2^{n}\bracks{\Gamma\pars{1/2 + n}/\Gamma\pars{1/2}}} = \overbrace{\root{\pi}}^{\ds{\Gamma\pars{1/2}}}\ \lim_{n \to \infty}{n! \over 2^{n}\pars{n - 1/2}!} \\[5mm] = &\ \root{\pi}\lim_{n \to \infty}{\root{2\pi}n^{n + 1/2}\expo{-n} \over 2^{n} \bracks{\root{2\pi}\pars{n - 1/2}^{n}\expo{-\pars{n - 1/2}}}} \\[5mm] = &\ \root{\pi} \lim_{n \to \infty}{n^{n + 1/2} \over 2^{n} \braces{n^{n}\bracks{1 - \pars{1/2}/n}^{\, n}\expo{1/2}}} = \root{\pi}\lim_{n \to \infty}{n^{1/2} \over 2^{n}} \\[5mm] = &\ \root{\pi}\lim_{n \to \infty}{1 \over 2^{n +1}\root{n}} = \bbx{0} \end{align}

Note that $\ds{\lim_{n \to \infty}\pars{1 - {1/2 \over n}}^{n} = \expo{-1/2}}$.