Prove $\lim\limits_{n \to \infty} \sup \left ( \frac{(2n - 1)^{2n - 1}}{2^{2n} (2n)!)} \right ) ^ {\frac 1 n} = \frac {e^2} 4$

Question

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.

Can you check for typos? Either the limit is $\frac{e^2}{2}$ or the expression is missing something. — Tom, Feb 20 '15 at 13:01
@Tom I checked the book, but it stays still e^2/4. It is surely possible that the book is not precise, too. — taro, Feb 21 '15 at 07:52
@Tom Indeed I found that I copied the formula wrong. There was 2^(2n) instead of 2^n. — taro, Feb 26 '15 at 19:12
@Tom I edited your answer, too, because I edited the question. — taro, Feb 26 '15 at 19:29

score 2 · Accepted Answer · edited Apr 13 '17 at 12:20

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 $\lim a_n^{1/n}=\frac{e^2}{4}$.

score 1 · Answer 2 · edited Feb 20 '15 at 12:24

1

Hint 1: $x = e^{\log x}$

Hint 2: $\log n! = \sum\limits_{k=1}^{n} \log k $

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

edited Feb 20 '15 at 12:24

rubik

9,344

answered Feb 20 '15 at 12:07

Alex

19,262

Prove $\lim\limits_{n \to \infty} \sup \left ( \frac{(2n - 1)^{2n - 1}}{2^{2n} (2n)!)} \right ) ^ {\frac 1 n} = \frac {e^2} 4$

2 Answers2