Stirling's Formula and Logarithms

Question

I'm going through the proof of Stirling's Formula and I'm not quite sure of the following line:

Since the log function is increasing on $(0,\infty)$, we get

$\int_{n-1}^n \log(x)dx < \log(n) < \int_n^{n+1} \log(x)dx$

The argument might be trivial, but I'm quite seeing it. I tried solving both the left and right hand sides of the inequalities and for the left side I get $$\log\frac{n^n}{(n-1)^{n-1}} - 1$$

I tried showing by induction that $\displaystyle\frac{n^n}{(n-1)^{n-1}} \le n$ but it fails for $n=2$. Any suggestions?

Answer 1

Hint: Use the mean-value theorem for integrals $\int_a^b f(x) dx = f(c)(b-a)$ along with the fact that $\log$ is increasing.

Answer 2

As suggested above you can write the following inequality:

\begin{equation} \log\frac{n^n}{(n-1)^{n-1}}-1<\log n\\\log\frac{n^n}{(n-1)^{n-1}}<\log n+1\\\log\frac{n^n}{(n-1)^{n-1}}<\log ne\\ \frac{n^n}{(n-1)^{n-1}}<ne \end{equation}

Now it is true the case $n=2$. However I think it is better following the way suggested in the comment above:$\int_{n-1}^n\log x dx<\int_{n-1}^n\log n dx=\log n<\int_n^{n+1}\log xdx$

Answer 3

To show $\int_{n-1}^n \log(x)dx < \log(n) < \int_n^{n+1} \log(x)dx $.

The basic result needed is this:

If $f(x)$ and $g(x)$ are continuous and $f(x) \ge g(x)$ on $[a, b]$, then $\int_a^b f(x) dx \ge \int_a^b g(x) dx $ with equality if and only if $f(x) = g(x)$ everywhere on $[a, b]$.

A corollary of this is that if $f(x)$ is strictly increasing on $[a, b]$ and $a < b$ then $f(a) < \dfrac1{b-a} \int_a^b f(x) dx < f(b) $.

All this says is that the average value of an increasing function is strictly between its min and max values.

The reverse inequality holds for strictly decreasing functions.

Since $\log(x)$ is increasing, $\log(n) \le \log(x) \le \log(n+1) $ for $n \le x \le n+1$ with equality only at one of the endpoints.

Therefore $\log (n) < \int_n^{n+1} \log(x) dx < \log(n+1) $.