TITLE: Almost Stirling's Approximation
QUESTION [21 upvotes]: I need to prove the following bound $$n! \le e \sqrt n \left( \frac n e \right)^n$$
I can bound $\ln 1 + \ln 2 + \dots + \ln n$ as a Riemann sum with the function $\ln(n+1)$ and the trapezoidal rule:
$$(\ln 1)/2 + \sum_{i=2}^n \ln i \ + (\ln(n+1))/2 < \int_0^n \ln (x+1) dx $$
Integrating I get the bound $$n! < \left( \frac{n+1}{e} \right)^n \sqrt{n+1}$$
As $n$ goes to infinity, my bound and the required bound get arbitrarily close (for $n=1$, the error is about $4\%$), but the one I need to prove is slightly tighter. How can I modify my bound to get the required bound?

REPLY [37 votes]: So you have
$n! 
\lt \left( \frac{n+1}{e} \right)^n \sqrt{n+1}
$.
I'll play and see what happens.
Replacing $n$ by $n-1$,
we get
$(n-1)! 
\lt \left( \frac{n}{e} \right)^{n-1} \sqrt{n}
$.
Multiplying by $n$,
we get
$\begin{array}\\
n! 
&\lt n\left( \frac{n}{e} \right)^{n-1} \sqrt{n}\\
&= e\left( \frac{n}{e} \right)^{n} \sqrt{n}\\
\end{array}
$
which is what you want!!!