$$n! \approx \sqrt{2 \pi n} \; \left(\frac{n}{\mathrm e}\right)^{n},$$ in the sense that the percentage error $\to 0$ as $n \to \infty$.
Show that the formula has an error of approximately $2.73\%$ for $3!$ and $0.83\%$ for $10!$.
Find the percentage error for $60!$.
I am assuming that there is an algebraic approach, that utilizes $3!$ and $10!$ to find the solution for $60!$, rather than substituting numbers.
To turn this function into one that outputs percentage error, I converted it to: $1 - \frac{\textrm{Stirlings formula}}{n!}$.
However, my calculator still is unable to compute it. I used an online calculator and managed to solve it correctly. The topic is on factorials, so I believe there is another way of solving it that utilized $3!$ and $10!$, but can’t find it. Is there another way? If so, how? Thanks!
I played with some numbers and I´ve got good approximations. Let
$$\mathcal E(n)=\frac{\sqrt{2 \pi n} \; \left(\frac{n}{\mathrm e}\right)^{n}-n!}{n!}$$
Then the approximation for relative errors of multiple n´s are
$$\mathcal E(n\cdot m)\approx\frac{\mathcal E(n)}{m}$$
For $n=10$ and several values of $m$ the results are
Remark
I´ve added a column with the relative differences (in percent). It shows that the numbers of that column are all far below $1\%$ (absolute value). I would call it a sufficiently accurate approximation for the most cases. A reply would be nice.