Is Ramanujan's approximation for the factorial optimal, or can it be tweaked? (answer below)

Ramanujan's famous factorial approximation, $$n!\approx\sqrt{\pi}\left(\frac{n}{e}\right)^n\root\LARGE{6}\of{8n^3+4n^2+n+\frac{1}{30}}$$ is far more accurate that the Stirling approximation when applied to numbers less than 1, but it's not quite there. What could be done to tweak it without adding to its complexity?

Update: After experimenting with Ramanujan's approximation, I've discovered this disturbing fact: apparently, the error decreases slower than the factorial increases, leading to massive integer-level errors the higher the factorial gets. Allow me to demonstrate: $$\begin{array}{r|rr} n&\text{actual }n!&\text{Ramanujan}\\\hline 0&1&1.005513858315898906334685\\ 1&1&1.000283346113497298280502\\ 2&2&2.000066137639113675155990\\ 3&6&6.000048293969899370824935\\ 4&24&24.000067662060676644042510\\ 5&120&120.000147065856635128019467\\ 6&720&720.000442402580258517644894\\ 7&5040&5040.001717876125295382828871\\ 8&40320&40320.008220460028928349902077\\ 9&362880&362880.046912269278701001557543\\ 10&3628800&3628800.311612606631103904381528\\ 11&39916800&39916802.364768173672637433190699\end{array}$$ At first the error reduction is apparent, but then the zeroes start to disappear, until at $11!$ the error overtakes the decimal point. Now I'm not saying the error is increasing—only that it it not decreasing fast enough. What can be done to correct this problem without radically altering the formula? I know that Ramanujan himself anticipated this problem when he wrote this inequality: $$\sqrt{\pi}\left(\frac{n}{e}\right)^n\root\LARGE{6}\of{8n^3+4n^2+n+\frac{1}{100}}<n!<\sqrt{\pi}\left(\frac{n}{e}\right)^n\root\LARGE{6}\of{8n^3+4n^2+n+\frac{1}{30}}$$ Could this hold the key? (See Formula 2.1 here)

Solution 1:

Interestingly, if we do a series expansion in Mathematica

Series[(Gamma[n + 1]/Sqrt[Pi] (n/Exp[1])^-n)^6, {n, Infinity, 6}]

We get $$8 n^3+4 n^2+n+\frac{1}{30}-\frac{11}{240 n}+\frac{79}{3360n^2}+\frac{3539}{201600 n^3}-\frac{9511}{403200 n^4}-\frac{10051}{716800n^5}+\frac{233934691}{6386688000n^6}+O\left(\frac{1}{n}\right)^{13/2}.$$ This suggests that your ad hoc adjustment of the constant term is probably not strictly justified.

Solution 2:

Long time ago, needing an explicit formula for statistical thermodynamics, I was facing the same problem of the calculation of $n!$ for $0<n<1$. Starting from Ramanujan's approximation, what I did was to write $$n!\approx\sqrt{\pi}\left(\frac{n}{e}\right)^n\root\LARGE{6}\of{8n^3+4n^2+n+x(n)}$$ What can be easily established is that $$x(0)=\frac{1}{\pi ^3}$$ just as proposed by Brian J. Fink. We can also show that $$x(1/2)=\frac{1}{8} \left(e^3-20\right)$$ and $$x(1)=13-\frac{e^6}{\pi ^3}$$ and I used (this is totally empirical and cannot be justified) a quadratic expansion of $x(n)=a+b n+c n^2$ which leads to $$a=\frac{1}{\pi ^3}$$ $$b=\frac{1}{2} \left(6+e^3-\frac{2 \left(3+e^6\right)}{\pi ^3}\right)$$ $$c=\frac{1}{2} \left(-32-e^3+\frac{4 \left(1+e^6\right)}{\pi ^3}\right)$$ Again, this has never been supposed to be used outside the range $[0,1]$