# Substituting large values of $n$ into Stirling’s formula, given the outcomes of other $n$ values

Mathematics Asked by Subbota on October 9, 2020

$$n! approx sqrt{2 pi n} ; left(frac{n}{mathrm e}right)^{n},$$
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

$$mathcal E(ncdot m)approxfrac{mathcal E(n)}{m}$$

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.

Correct answer by callculus on October 9, 2020