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.

Hint 1: $x  = e^{\log x}$
Hint 2: $\log n! = \sum\limits_{k=1}^{n} \log k $
Hint 3: $\int_{1}^{n} \log x dx < \sum\limits_{k=1}^{n} \log k < \int_{1}^{n+1} \log x dx$
