Use trapezoidal rule to find $\lim _{n\rightarrow \infty }\frac{2^{2n}e^{-n}n^{n}n!}{\left( 2n\right) !}$ 
When $f\left( x\right) =\log x$ and section $\left[ 1,2\right]$
$$\lim _{n\rightarrow \infty }n\left[ \int _{a}^{b}f\left( x\right) dx-\dfrac{b-a}{n} \left\{\dfrac{f\left( a\right) +f\left( b\right) }{2}+\sum ^{n-1}_{k=1}f\left( a+\dfrac{\left( b-a\right) k}{n}\right) \right\} \right] =0$$
Use this to find the value of the limit:
$$\lim _{n\rightarrow \infty }\dfrac{2^{2n}e^{-n}n^{n}n!}{\left( 2n\right) !}-①$$

I have assigned $f(x)$ and section $\left[ 1,2\right]$.
$$\lim _{n\rightarrow \infty }n\left[ \dfrac{1}{2}\log 2-\dfrac{1}{n}\left\{\dfrac{1}{2}\log 2+\sum ^{n-1}_{k=1}\log \left( 1+\dfrac{k}{n}\right) \right\} \right]-②$$
I can't find common points of $①$ and $②$. So I don't know what to do next.
 A: This answer is incorrect.
I'm leaving it here (instead of deleting it, got the advice here) just so that other solvers don't make the same mistake.
As rightly pointed out in the comments, This lemma should be used to obtain the correct answer of $1/\sqrt 2$

$$\begin{gather}
L = \lim_{n \to \infty} \left( \frac{4}{e} \right)^n \cdot \left( \frac{n}{n+1} \cdot \frac{n}{n+2} ... \frac{n}{n+n}\right) \\
\log L = \lim_{n \to \infty} n \log \left( \frac 4e \right) + \sum_{r=1}^n\log\left(\frac{n}{n+r}\right) \\
\log L = \lim_{n \to \infty} n \left( \log \left( \frac 4e \right) - \frac 1n\sum_{r=1}^{n}\log\left(1+\frac rn\right)\right) 
\end{gather}$$
the term on the right is a Riemann sum, and it can be derived by using the trapezoidal rule to get the area for infinitesimal parts and summing them up. I'll leave this part up to you but the solution of the riemann sum is $\log\left( \frac 4e\right)$. This makes the limit
$$\log L = \lim_{n \to \infty} n \left( \log \left( \frac 4e \right) - \log \left( \frac 4e \right) \right) \\
\log L = 0 \\
\boxed{L = 1}$$
A: $$a_n=\dfrac{2^{2n}e^{-n}n^{n}n!}{\left( 2n\right) !}\implies \log(a_n)=2n \log(2)-n+n \log(n)+\log(n!)-\log((2n)!)$$
Use Stirling approximation twice and continue with Taylor expansion. Yous should obtain
$$\log(a_n)=-\frac{\log (2)}{2}+\frac{1}{24 n}+O\left(\frac{1}{n^3}\right)$$ Just continue
$$a_n=e^{\log(a_n)}=\frac{1}{\sqrt{2}}\left(1+\frac{1}{24 n}+\frac{1}{1152 n^2}\right)+O\left(\frac{1}{n^3}\right)$$
