Problem 2 - Stirling's Formula

For an integer $n\geq 2$, define $a_n$ by

$$a_n = \sum_{k=1}^{n-1}\log k + \frac{1}{2}\log n$$

where $\log$ is the natural logarithm and $e$ below is the base of the natural logarithm.

1.

Show $a_n < \int_1^n \log xdx$.

2.

Show $\int_{\frac{3}{2}}^n \log xdx < a_n$.

3.

Setting $\delta_n = 1 + a_n - \int_1^n \log x dx$, show that, when $n\rightarrow\infty$, $\delta_n$ converges to a constant.

4.

Denoting the converging value of $\delta_n$ by $\delta$, show

$$\lim_{n\rightarrow\infty} \frac{n!}{e^\delta \sqrt{n}\left(\frac{n}{e}\right)^n}$$

exists, and compute its value.

1.

$\forall x>0, \frac{d^2\log x}{dx^2} = -\frac{1}{x^2} < 0$, that is $:x\mapsto\log x$ is concave.

We thus have the following inequality:

$$\int_{k}^{k+1} \log x dx > \frac{1}{2}(\log k + \log k+1)$$

From which we deduce the sequence below:
$\int_1^2 \log x dx > \frac{1}{2}(\log 1 + \log 2)$
$\int_2^3 \log x dx > \frac{1}{2}(\log 2 + \log 3)$
$\int_3^4 \log x dx > \frac{1}{2}(\log 3 + \log 4)$
$\vdots$
$\int_{n-2}^{n-1} \log x dx > \frac{1}{2}(\log (n-2) + \log (n-1))$
$\int_{n-1}^n \log x dx > \frac{1}{2}(\log (n-1) + \log (n))$
And therefore:

$$\int_1^n \log x dx > \sum_{k=2}^n \log k + \frac{1}{2}(\log 1 + \log n) = a_n$$

2.

$\forall x>0, \frac{d^2\log x}{dx^2} = -\frac{1}{x^2} < 0$, that is $:x\mapsto\log x$ is concave.

Thus, the area denoted 1 is larger than the one denoted 2 on the previous figure. We thus have the following inequality:

$$\int_{k-\frac{1}{2}}^{k+\frac{1}{2}} \log x dx < \log k$$

From which we deduce the sequence below:
$\int_{\frac{3}{2}}^{\frac{5}{2}} \log x dx < \log 2$
$\int_{\frac{3}{2}}^{\frac{7}{2}} \log x dx < \log 3$
$\vdots$
$\int_{n-\frac{3}{2}}^{n-\frac{1}{2}} \log x dx < \log (n-1)$
and
$\int_{n-\frac{1}{2}}^n \log x dx < \frac{1}{2} \log n$
And therefore:

$$\int_{\frac{3}{2}}^n \log x dx < \sum_{k=2}^{n-1} \log k + \frac{1}{2} \log n = a_n$$

3.

$$\forall n \geq 2, \int_{\frac{3}{2}}^n \log x dx < a_n < \int_1^n \log x dx$$

$$1 - \int_1^\frac{3}{2} \log x dx < \delta_n < 1$$

Thus, $\delta_n$ has upper and lower limits. Let us determine whether $\delta_n$ is increasing or decreasing:
$\delta_{n+1} - \delta_n = a_{n+1} - a_{n} - \int_1^{n+1}\log xdx + \int_1^n \log xdx$
$= \log (n) + \frac{1}{2}\log(n+1) - \frac{1}{2}\log n - \int_{n}^{n+1} \log x dx$
$= \frac{1}{2}\left( \log (n) + \log (n+1) \right) - \int_n^{n+1} \log x dx$
$<0$ as we saw above.

This means that $\delta_n$ is decreasing with a lower limit, thus it converges to a constant we shall call $\delta = \lim_{\infty} \delta_n$.

4.

$\delta_n = 1 + \sum_{k=1}^{n-1}\log k + \frac{1}{2}\log n - \int_1^n \log xdx$
$= 1 + \log n! -\frac{1}{2}\log n - \left[ x(\log x - 1)\right]_1^n$
$= 1 + \log n! - \frac{1}{2}\log n - n(\log n - 1) - 1$
$= \log n! - \left( n + \frac{1}{2}\right)\log n + n$
But, $\delta_n\rightarrow_{\infty}\delta$. Which means that:
$\log n! - \left( n + \frac{1}{2}\right)\log n + n \rightarrow_\infty \delta$
$n! n^{-n-\frac{1}{2}} e^n \rightarrow_\infty e^\delta$
$\frac{n!}{e^\delta \sqrt(n) \left( \frac{n}{e} \right)^n} \rightarrow_\infty 1$