Double Factorial, Double Headache!!

The double factorials can be written as follows:

$\displaystyle (2n)!!=2^n\cdot n!$

$\displaystyle (2n-1)!!=\frac{(2n)!}{2^n\cdot n!}$

(Some info about double factorial here: Double Factorial - Wolfram Mathworld )

Hence, the given limit is:

$\displaystyle \lim_{n\rightarrow \infty} \frac{2^{2n}(n!)^2}{\sqrt{n}(2n)!}$

Apply Stirling's approximation to obtain:

$\displaystyle \lim_{n\rightarrow \infty} \frac{2^{2n} (2\pi n)(n/e)^{2n}}{\sqrt{n}\sqrt{2\pi (2n)}(2n/e)^{2n}}$

All the n's cancel and we are left with:

$\displaystyle \lim_{n\rightarrow \infty} \frac{2\pi}{\sqrt{4\pi}}=\boxed{\sqrt{\pi}}$

If you take the square of the limit, then it's simply twice the value of Wallis Product , so the answer is simply $\boxed{ \sqrt{\pi} }$

We use the fact that $(2n-1)!!=\frac{(2n)!}{2^n \cdot n!}$ (Info regarding this fact can be found here: Double Factorials )

Now we show that $(2n)!!=n! \cdot 2^n$ :

$(2n)!! =\frac{(2n)!}{(2n-1)!!}$ $\large =\frac{(2n)!}{\frac{(2n)!}{2^n \cdot n!}}$ $=2^n \cdot n!$ . We now substitute these expressions in to the limit to get:

$\displaystyle \lim_{n \rightarrow \infty}\frac{2^n \cdot n!}{\sqrt{n} \cdot \frac{(2n)!}{2^n \cdot n!}}=\lim_{n \rightarrow \infty} \frac{(2^n \cdot n!)^2}{\sqrt{n} \cdot (2n)!}$

Since our limit's expression is now entirely in terms of "single" factorials, it would be wise to make use of Stirling's Approximation . The approximation suggests that $n! \sim \sqrt{2 \pi n} \left(\frac{n}{e}\right) ^n$ . It should also be noted that the error term in the approximation term becomes very small and the approximation becomes very accurate as $n \rightarrow \infty$ , which is quite reasonable in the case of our limit. By the approximation we have that $(2n)! \sim \sqrt{4 \pi n} \left(\frac{2n}{e}\right) ^{2n}$ . Once again, making these substitutions in to our limit yields:

$\displaystyle \lim_{n \rightarrow \infty} \frac{\left(2^n \cdot \sqrt{2 \pi n}\left(\frac{n}{e}\right) ^n\right)^2}{\sqrt{n} \cdot \sqrt{4 \pi n} \left(\frac{2n}{e}\right) ^{2n} }$ $\displaystyle = \lim_{n \rightarrow \infty} \frac{2^{2n} \cdot 2 \pi n \left(\frac{n}{e}\right)^{2n}}{\sqrt{n} \cdot \sqrt{4 \pi n} \cdot 2^{2n} \left(\frac{n}{e}\right)^{2n}} = \lim_{n \rightarrow \infty} \frac{ 2 \pi n}{\sqrt{n} \cdot \sqrt{4 \pi n}}$

$\displaystyle =\frac{2 \pi}{2\sqrt{\pi}} \lim_{n \rightarrow \infty}(1) =\sqrt{\pi} \approx \boxed{1.772}.$

And we are done =D.

Using "stirling Aproximation"