I have developed the following limit $$L=\lim_{n\to\infty} \left(\frac{e}{4n + 2}\right)^n \frac{(2n)!}{n!}$$ Numerical tests have revealed $L\approxeq 0.859$.
I have also attempted to view the problem as $$L=\lim_{n\to\infty}\left(\frac{e}{2}\right)^n \prod_{i=1}^{n} \frac{n + i}{2n + 1}$$
Question: How would I prove that $L$ is a finite, nonzero number.
Use logarithm on expression under limit to get (after some effort) $$n-n\log(4+2/n)+\sum_{k=1}^{n}\log(1+k/n)\tag{1}$$ Next use the result that if $f\in C^1[0,1]$ then $$\sum_{k=1}^{n}f(k/n)-n\int_0^1 f(x) \, dx\to\frac{f(1)-f(0)}{2}$$ with $f(x) =\log(1+x)$ to get $$\sum_{k=1}^{n}\log(1+k/n)-n\int_0^1\log(1+x)\,dx\to \frac{1}{2}\log 2$$ ie $$\sum_{k=1}^{n}\log(1+k/n)-n\log 4+n\to\frac{1}{2}\log 2\tag{2}$$ The expression in $(1)$ can be rewritten as $$\sum_{k=1}^n\log(1+k/n)-n\log 4+n-n\log(1+1/2n)$$ and by $(2)$ this tends to $(1/2)\log 2-(1/2)$. The desired limit is thus $\sqrt{2/e}$.
