A Riemann sum III

Using Riemann sums evaluate the limit
$\ell = \lim_{n \rightarrow +\infty} \left ( \frac{1}{n} \ln n! - \ln n \right )$
Using the above result prove that
$\lim_{n \rightarrow +\infty} \left ( \frac{n! e^n}{n^{n+1/2}\sqrt{2\pi}} \right )^{1/n} =1$

Solution

We have successively:
$\begin{align*} \ell &= \lim_{n \rightarrow +\infty} \left ( \frac{1}{n} \ln n! - \ln n \right ) \\ &= \lim_{n \rightarrow +\infty} \left ( \frac{1}{n} \sum_{k=1}^{n} \ln k - \ln n \right ) \\ &= \lim_{n \rightarrow +\infty} \frac{1}{n} \sum_{k=1}^{n} \ln \frac{k}{n} \\ &= \int_{0}^{1} \ln x \, \mathrm{d}x \\ &= -1 \end{align*}$
Let $\displaystyle a_n = \left ( \frac{n! e^n}{n^{n+1/2}\sqrt{2\pi}} \right )^{1/n}$ . Taking logarithms on both sides we get that
$\begin{align*} \ln a_n &= \frac{1}{n} \left ( \ln n! + n \ln e - \left ( n + \frac{1}{2} \right ) \ln n - \frac{\ln 2 \pi}{2} \right ) \\ &= \frac{1}{n} \ln n! +1 - \ln n - \frac{\ln n}{2n} - \frac{\ln 2 \pi}{2n}\\ &=\left (\frac{1}{n} \ln n! - \ln n \right )+ 1 - \left ( \frac{\ln n}{n} + \frac{\ln 2\pi}{2n} \right ) \\ &\rightarrow -1 +1 -0 =0 \end{align*}$

The result follows.

A Fibonacci series