The task is to find $$L=\lim_{n\rightarrow \infty} \frac{1}{2n}\sqrt[n]{\frac{(2n+1)!}{n!}}.$$
I say we resort to Sirling's approximation $k!\approx \sqrt{2\pi k} (\frac{k}{e})^k$:
$$ L \approx \lim_n \frac{1}{2n}\frac{\bigg(\sqrt{2\pi (2n+1)}(\frac{2n+1}{e})^{2n+1}\bigg)^{1/n}}{\bigg(\sqrt{2\pi n}(\frac{n}{e})^{n}\bigg)^{1/n}}\approx \frac {4n^2/e^2}{2n^2/e} \approx \frac {2}{e}.$$
however, as usual there is a solution manual that disagrees and says $L=2$. I'd appreciate it if you could help me figure out what's wrong.
You can take logs to get the expression $$\frac{1}{n}\sum_{i=0}^{n}\log(n+1+i)-\log(2n)$$ whose limit is same as that of $$\frac {1}{n}\sum_{i=1}^{n}\log(n+i)-\log (2n)=\frac{1}{n}\sum_{i=1}^{n}\log(1+i/n)-\log 2$$ and you can see the limit as $$\int_{0}^{1}\log(1+x)\,dx-\log 2=\log 2 - 1$$ and thus $L=2/e$.
The approach avoids the Stirling approximation and one needs a little care when using Stirling.
You may also try the standard result that if $a_n>0$ and $a_{n+1}/a_n\to L$ then $a_n^{1/n}\to L$. Here use $a_n=b_n^n$ where $b_n$ is the expression under limit in question.