Using Wallis's formula, show that ((2n-1)!!)/((2n)!!)
Using Wallis's formula, show that $\displaystyle\frac{(2n-1)!!}{(2n)!!}\sim\frac1{\sqrt{\pi n}}$ as $n\to\infty$.
Explanation of Method
Wallis's product (Problem 372), rewritten via double factorials, gives $\bigl[(2n)!!\bigr]^2/\bigl[(2n-1)!!(2n+1)!!\bigr]\to\pi/2$. Using $(2n+1)!!=(2n+1)(2n-1)!!$ turns this into a clean statement about the ratio $(2n)!!/(2n-1)!!$ alone, which rearranges directly to the target asymptotic.
- Rewrite Wallis's product in double-factorial form.
- Isolate the ratio $(2n)!!/(2n-1)!!$ using $(2n+1)!!=(2n+1)(2n-1)!!$.
- Solve asymptotically and take reciprocals.
Step 1 · Wallis's product in double-factorial form
From Problem 372, $\prod_{n=1}^\infty\frac{(2n)^2}{(2n-1)(2n+1)}=\frac\pi2$, which is equivalent to $$ \lim_{n\to\infty}\frac{\bigl[(2n)!!\bigr]^2}{(2n-1)!!\,(2n+1)!!}=\frac\pi2. $$
Step 2 · Isolate the ratio
Since $(2n+1)!!=(2n+1)(2n-1)!!$: $$ \frac{\bigl[(2n)!!\bigr]^2}{\bigl[(2n-1)!!\bigr]^2(2n+1)}\longrightarrow\frac\pi2 \implies \left[\frac{(2n)!!}{(2n-1)!!}\right]^2 \sim \frac{\pi(2n+1)}2 \sim \pi n. $$
Step 3 · Solve and take the reciprocal
Taking square roots: $$ \frac{(2n)!!}{(2n-1)!!}\sim\sqrt{\pi n} \implies \boxed{\frac{(2n-1)!!}{(2n)!!}\sim\frac1{\sqrt{\pi n}}} $$ (Numerical check: at $n=10^4$, the true ratio is $\approx0.0056418253$ versus $1/\sqrt{\pi\cdot10^4}\approx0.0056418958$ — matching to 4 significant figures, with the ratio of the two $\to1$ as $n\to\infty$, e.g. $0.99999$ at $n=10^4$.)
Summary Table
| Property | Value |
|---|---|
| Wallis (double-factorial form) | $\bigl[(2n)!!\bigr]^2/\bigl[(2n-1)!!(2n+1)!!\bigr]\to\pi/2$ |
| Asymptotic ratio | $(2n-1)!!/(2n)!!\sim1/\sqrt{\pi n}$ |
Additional Notes
- This asymptotic can be independently cross-checked via Stirling's formula: writing $(2n-1)!!/(2n)!!=(2n)!/[2^{2n}(n!)^2]$ (the central binomial coefficient's reciprocal, up to a factor of $\binom{2n}n$) and applying $n!\sim\sqrt{2\pi n}(n/e)^n$ reproduces exactly $1/\sqrt{\pi n}$, confirming the Wallis-based derivation independently.
- This is precisely the normalizing constant that appears in the de Moivre–Laplace / central limit theorem approximation to the binomial distribution at its center: $\binom{2n}n2^{-2n}\sim1/\sqrt{\pi n}$, linking Wallis's 1655 product to 18th-century probability theory.
- The rate of convergence in Step 3 is $O(1/n)$ inside the bracket (from $2n+1\sim2n$), which becomes $O(1/n)$ relative error in the ratio itself after the square root — the same slow Wallis-characteristic convergence rate seen throughout Problem 372.