$\pi$ is everywhere!

Double factorials can be defined as follows in terms of regular factorials:

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

Therefore, the limit simplifies to: $\lim_{n\to\infty} \frac{(2n)!\sqrt{n}}{(n!)^2 2^{2n}}$

Now apply Stirling's Approximation: $\lim_{n\to\infty} \frac{(2n)!\sqrt{n}}{(n!)^2 2^{2n}} = \lim_{n\to\infty} \frac{(2n)^{2n}e^{-2n}\sqrt{4\pi n}}{(n^n e^{-n}\sqrt{2\pi n})^2 2^{2n}} \sqrt{n} = \frac{1}{\sqrt{\pi}} = \pi^{-1/2}$ Therefore the answer is $1 + 2 = \boxed{3}.$