Prove $\lim\limits_{n \to \infty} \sup \left ( \frac{(2n - 1)^{2n - 1}}{2^{2n} (2n)!)} \right ) ^ {\frac 1 n} = \frac {e^2} 4$ This is a problem in Heuer (2009) "Lerbuch der Analysis Teil 1" on page 366. I assume that the proof should use $e = \sum\limits_{k = 0}^{\infty} \frac 1 {k!}$, but I cannot come further.

If $a_n=\frac{(2n - 1)^{2n - 1}}{2^{2n} (2n)!}$. Compute the limit
$$\begin{align}\lim \frac{a_{n+1}}{a_n}&=\lim \frac{\frac{(2n + 1)^{2n + 1}}{2^{2n+2} (2n+2)!}}{\frac{(2n - 1)^{2n - 1}}{2^{2n} (2n)!}}\\&=\lim\frac{(2n+1)(2n+1)}{4(2n+1)(2n+2)}\left(\frac{2n+1}{2n-1}\right)^{2n-1}\\&=\frac{1}{4}\lim\left[\left(1+\frac{2}{2n-1}\right)^{\frac{2n-1}{2}}\right]^2\\&=\frac{1}{4}e^2\end{align}$$
Since $\lim\frac{a_{n+1}}{a_n}=\frac{e^2}{4}$ it follows that $\lim a_n^{1/n}=\frac{e^2}{4}$.
