Continued Fractional multiplication

It is just equal to $\displaystyle {lim}_{n->\infty}(\frac{(2n)!!^2}{(2n+1)!!(2n-1)!!})$

Now, using stirling's approximation(of course we have to simplify this double factorial into single one and so on but that's just trivial),

it equals -

$\displaystyle {lim}_{n->\infty}(\frac{n\pi}{2n+1})$

which is just(you can use anything, L'hopital's rule preferably)

$\displaystyle \boxed{\frac{\pi}{2}}$

We need to find $\lim_{n\to\infty}\frac{(n!!)^2}{(n+1)!!(n-1)!!}$ for even $n$ . We will use some basic results from Calculus I. Let $a_n = \int\limits_{0}^{\pi/2}\sin^{n}x dx$ . Using Integration by parts, one can show that $a_n= \frac{(n-1)!!}{n!!}$ for odd $n$ and $a_n= \frac{(n-1)!!}{n!!}\frac{\pi}{2}$ for even $n$ . Also, it is easy to see that $\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=1$ . For even $n$ , we have $\frac{a_{n+1}}{a_n}=\frac{(n!!)^2}{(n+1)!!(n-1)!!}\frac{2}{\pi}$ . Taking the limit and multiplying with $\frac{\pi}{2}$ , we find that $\lim_{n\to\infty}\frac{(n!!)^2}{(n+1)!!(n-1)!!}=\frac{\pi}{2}.$

If you view the image, you see that it is literally half a pie. Therefore, logically, we may assume that answer

This series reduces to 2 times the convergence value for the series product of (n^2-1)/n^2, where n=3,5,7,... A simple computer program shows that the series product converges on pi/4, so 2 times this is pi/2.