It is proved in Advanced Calculus by Angus Taylor, ยง 20.8, that
$$\log n!=\log \left( \left( \frac{n}{e}\right) ^{n}\sqrt{2\pi n}\right) +r_{n},$$
where
$$r_{n}=\sum_{k=1}^{\infty }S_{k}$$
with
$$S_{k}=\sum_{p=n+1}^{\infty }\frac{k}{2(k+1)(k+2)p^{k+1}}.$$
This formula for $r_{n}$ provides a method for finding the estimate
$$\frac{1}{12\left( n+1\right) } Indeed $$\frac{1}{k\left( n+1\right) ^{k}}=\sum_{p=n+1}^{\infty }\int_{p}^{p+1}\frac{%
1}{x^{k+1}}\mathrm dx<\sum_{p=n+1}^{\infty }\frac{1}{p^{k+1}}$$ $$\sum_{p=n+1}^{\infty }\frac{1}{p^{k+1}}<\sum_{p=n+1}^{\infty }\int_{p-1}^{p}%
\frac{1}{x^{k+1}}\mathrm dx=\frac{1}{kn^{k}},$$ and so $$\frac{1}{2(k+1)(k+2)\left( n+1\right) ^{k}} $$\frac{1}{12\left( n+1\right) }<\sum_{k=1}^{\infty }\frac{1}{%
2(k+1)(k+2)\left( n+1\right) ^{k}} Added:The Wikipedia article pointed to in the comment gives an approximation based on the Euler-MacLaurin formula in terms of the Bernoulli numbers and states that it can also be obtained by repeated integration by parts. The Stirling series that appears in the same article is different from the one above. Question (edited): Are there better estimates of $r_{n}$ based on different methods?