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

Question

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

callculus · Accepted Answer

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.

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

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