How can I calculate $\lim_{n \rightarrow \infty} \frac {e^n(2n)!}{(4n)^nn!}$ How should I get the result $$\lim_{n \rightarrow \infty} \frac {e^n(2n)!}{(4n)^nn!}=\sqrt2$$
without relying on Stirling's formula or the Central Limit Theorem?
I am totally clueless. Can somebody help me?
 A: This is based on the following lemma.

Lemma: If $f\in C^{1}[0,1] $ then $$\lim_{n\to\infty} \sum_{k=1}^{n}f\left(\frac{k}{n}\right)-n\int_{0}^{1}f(x)\,dx=\frac{f(1)-f(0)}{2}$$
  (proof available here) 

In the above lemma it is sufficient to assume the Riemann integrability of $f'$. 
If $a_n$ is the sequence in question and $L$ is the desired limit then we have
\begin{align} 
\log L&=\log \lim_{n\to\infty}a_n\notag\\
&=\lim_{n\to \infty} \log a_n\notag \\
&=\lim_{n\to\infty} n\log\left(\frac{e}{4}\right)+\sum_{k=1}^{n}\log\left(1+\frac{k}{n}\right)\notag\\
&=\lim_{n\to\infty} \sum_{k=1}^{n}f\left(\frac{k}{n}\right)-n\int_{0}^{1}f(x)\,dx\notag
\end{align}
where $f(x) =\log(1+x)$. Using the lemma we see that $$\log L=\frac{\log 2}{2}$$ and hence $L=\sqrt{2}$.
A: Solution
According to Stirling's formula, $$n!=\sqrt{2\pi n} \left(\frac{n}{e}\right)^n \cdot e^{\frac{\theta}{12n}}~~,(0<\theta<1)$$
Hence, $$\lim_{n \rightarrow \infty} \frac {e^n(2n)!}{(4n)^nn!}=\lim_{n \to \infty}\frac{e^n\sqrt{4\pi n} \left(\frac{2n}{e}\right)^{2n} \cdot e^{\frac{\theta_1}{24n}}}{(4n)^n \sqrt{2\pi n} \left(\frac{n}{e}\right)^n \cdot e^{\frac{\theta_2}{12n}}}=\sqrt{2}.$$
