Asymptotic formula for ratio of double factorials

Question

I am interested in getting an asymptotic formula for $$ \frac{(2n-1)!!} {(2n)!!}. $$

After some trial and error, I think the asymptotic can be of the form $\frac{1}{\sqrt{\alpha n + \beta}}.$

Use Stirling's formula, I got $\alpha = \pi$

Using software, I was able to get $\beta = \frac{\pi}{4}$, but I have no idea how to prove this.

So my question is: how do I show that $$ \lim_{n \to \infty} \left(\left(\frac{(2n)!!}{(2n-1)!!}\right)^{2} - \pi n\right) = \frac{\pi}{4} $$ ?

Answer 1

If you take an extra term in Stirling's approximation, you have

• $n! = \sqrt{2\pi n}\left(\frac{n}{e}\right)^n \left(1 +\frac{1}{12n} + O(\frac1{n^2})\right)$
• $(2n)! =\sqrt{4\pi n}\left(\frac{2n}{e}\right)^{2n} \left(1 +\frac{1}{24n} + O(\frac1{n^2})\right)$
• $(2n)!! = 2^n n! = \sqrt{2n\pi}\left(\frac{2n}{e}\right)^{n} \left(1 +\frac{1}{12n} + O(\frac1{n^2})\right)$
• $(2n-1)!! = \frac{(2n)!}{(2n)!!} = \sqrt{2}\left(\frac{2n}{e}\right)^{n} \left(1 -\frac{1}{24n} + O(\frac1{n^2})\right)$
• $\frac{(2n)!!}{(2n-1)!!} = \sqrt{n\pi}\left(1 +\frac{1}{8n} + O(\frac1{n^2})\right)$
• $\left(\frac{(2n)!!}{(2n-1)!!} \right)^2= {n\pi}\left(1 +\frac{1}{4n} + O(\frac1{n^2})\right) = n\pi+\frac\pi4+ O(\frac1{n})$

which seems to be what you want.

Answer 2

$$A_n=\frac{(2n-1)!!} {(2n)!!}=\frac{\Gamma \left(n+\frac{1}{2}\right)}{\sqrt{\pi } \,\,\Gamma (n+1)}$$ Take logarithms, use twice Stirling approximation and then Taylor series $$\log(A_n)=-\frac{1}{2} (\log (n)+\log (\pi ))-\frac{1}{8 n}+\frac{1}{192 n^3}+O\left(\frac{1}{n^5}\right)$$ Continuing wtit Taylor $$A_n=e^{\log(A_n)}=\frac{1}{\sqrt{n\pi }} \left(1-\frac{1}{8 n}+\frac{1}{128 n^2}+\frac{5}{1024 n^3}+O\left(\frac{1}{n^4}\right) \right)$$ $$\frac 1{A_n^2}-\pi n=\frac \pi 4\left(1+\frac{1}{8 n}-\frac{1}{32 n^2}+O\left(\frac{1}{n^3}\right) \right)$$ which is in a relative error of $0.01$% when $n>4$.

Answer 3

Note that \begin{align*} \frac{{(2n - 1)!!}}{{(2n)!!}} & = \frac{{\Gamma \!\left( {n + \frac{1}{2}} \right)}}{{\sqrt \pi \Gamma (n + 1)}} = \frac{1}{\pi }\int_0^1 {t^{n - 1/2} (1 - t)^{ - 1/2} {\rm d}t} \\ & = \frac{1}{\pi }\int_0^{ + \infty } {{\rm e}^{ - (n + 1/2)s} (1 - {\rm e}^{ - s} )^{ - 1/2} {\rm d}s} \\ & = \frac{1}{\pi }\int_0^{ + \infty } {{\rm e}^{ - (n + 1/4)s} s^{ - 1/2} \sqrt {\frac{{s/2}}{{\sinh (s/2)}}} {\rm d}s} \end{align*} for $n\ge 0$. Now $$ s^{ - 1/2} \sqrt {\frac{{s/2}}{{\sinh (s/2)}}} = s^{ - 1/2} - \frac{{s^{3/2} }}{{48}} + \frac{{s^{7/2} }}{{2560}} - \ldots $$ as $s\to 0^+$. Thus, by Watson's lemma, $$ \frac{{(2n - 1)!!}}{{(2n)!!}} \sim \frac{1}{{\sqrt {\pi (n + 1/4)} }}\left( {1 - \frac{1}{{64(n + 1/4)^2 }} + \frac{3}{{4096(n + 1/4)^4 }} - \ldots } \right) $$ as $n\to+\infty$.

Answer 4

Inequality $(9)$ in this answer says that $$ \frac{4^n}{\sqrt{\pi\!\left(n+\frac14+\frac1{16n+12}\right)}}\le\binom{2n}{n}\le\frac{4^n}{\sqrt{\pi\!\left(n+\frac14\right)}}\tag1 $$ Therefore, $$ \begin{align} \frac{\color{#C00}{(2n)!!}}{\color{#090}{(2n-1)!!}} &=\frac{\color{#C00}{2^nn!}\color{#090}{2^nn!}}{\color{#090}{(2n)!}}\tag{2a}\\[3pt] &=\frac{4^n}{\binom{2n}{n}}\tag{2b}\\ &\in\left[\sqrt{\pi\!\left(n+\tfrac14\right)},\sqrt{\pi\!\left(n+\tfrac14+\tfrac1{16n+12}\right)}\right]\tag{2c} \end{align} $$ Explanation:
$\text{(2a):}$ rewrite double factorials in terms of factorials
$\text{(2b):}$ $\binom{2n}{n}=\frac{(2n)!}{n!n!}$
$\text{(2c):}$ apply $(1)$

Squaring $(2)$ and subtracting $\pi n$ gives $$ \frac\pi4\le\left(\frac{(2n)!!}{(2n-1)!!}\right)^2-\pi n\le\frac\pi4+\frac\pi{16n+12}\tag3 $$ Evaluate the limit with the Squeeze Theorem.