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

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.

The fraction $\tfrac{1}{30}$ may not be, practically speaking, the best upper bound for the final term of the radicand. Indeed, according to Ramanujan's own notes, the number ranges from $\tfrac{1}{30}$ down to $\tfrac{1}{100}$. Observe that when evaluated at $0$, $$\begin{align}0!&\approx\sqrt{\pi}\left(\frac{0}{e}\right)^0\root{\LARGE{6}}\of{8\cdot0^3+4\cdot0^2+0+\frac{1}{30}}\\ &=\sqrt{\pi}\cdot1\cdot\root{\LARGE{6}}\of{\frac{1}{30}}\\ &=\frac{\sqrt{\pi}}{\root{\LARGE{6}}\of{30}}\\ &=\root{\LARGE{6}}\of{\frac{\pi^3}{30}}\\ &\approx1.00551\end{align}$$ Notice the error in the result. However, if we replace the number $30$ in the above formula with $\pi^3$, the result is dead accurate: $$\begin{align}0!&\approx\sqrt{\pi}\left(\frac{0}{e}\right)^0\root{\LARGE{6}}\of{8\cdot0^3+4\cdot0^2+0+\frac{1}{\pi^3}}\\ &=\sqrt{\pi}\cdot1\cdot\root{\LARGE{6}}\of{\frac{1}{\pi^3}}\\ &=\frac{\sqrt{\pi}}{\root{\LARGE{6}}\of{\pi^3}}\\ &=\root{\LARGE{6}}\of{\frac{\pi^3}{\pi^3}}\\ &=1\end{align}$$ Now let's try it on $n=1$: $$\begin{align}1!&\approx\sqrt{\pi}\left(\frac{1}{e}\right)^1\root{\LARGE{6}}\of{8\cdot1^3+4\cdot1^2+1+\frac{1}{\pi^3}}\\ &=\frac{\sqrt{\pi}}{e}\root{\LARGE{6}}\of{13+\frac{1}{\pi^3}}\\ &\approx1.0002695\end{align}$$ For this reason, I recommend this modification to Ramanujan's approximation: $$n!\approx\sqrt{\pi}\left(\frac{n}{e}\right)^n\root\LARGE{6}\of{8n^3+4n^2+n+\frac{1}{\pi^3}}$$ Alternatively, since $\pi^3\approx31.00627668$, a slightly less robust adjustment to the formula can be made by substituting $31$ for $30$, instead of $\pi^3$.

Let me know what you think about this!

Additional thoughts: Based on the material I added to my question above, perhaps $\tfrac{1}{\pi^3}\approx31$ should be gradually reduced toward $\tfrac{1}{\pi^4}\approx97$ as $n$ increases. A draft formula, pending further investigation: $${\large n!\approx\sqrt{\pi}\left(\frac{n}{e}\right)^n\root\LARGE{6}\of{8n^3+4n^2+n+\frac{1}{\pi^{3+c}}}}$$ where $$\left(\frac{n}{n+1}\right)^3>c>\left(\frac{n}{n+1}\right)^2$$ The formula above is constantly evolving as I test and make adjustments. This formula is tricky to adjust. I have to express it in terms of an unknown value $c$ until I can dial it in, so to speak. My current estimate is $$c\approx\left(\frac{n}{n+1}\right)^{{\pi^3}\lower{2pt}/\lower{4pt}{12}}$$ Conclusion: The answer is no and no. According to my own analysis, Ramanujan's formula is only useful for estimating what the value of $n!$ might be for some $n$; and the error reduction rate appears painfully slow, slow enough to begin to accumulate error in the integer range as early as $11!$; as for tweaking it, the need appears to be for an adjustment of a major sort, which perhaps I will be unable to accomplish at this time. Indeed, any adjustment that optimizes the formula over one range appears to diminish its accuracy over another, as with all asymptotic approximations of Factorial and Gamma functions in general. I'm not saying a formula for the factorial function cannot be found that is optimal for all $x\geqslant0$; I simply have neither the time nor the mathematical expertise to do it myself.

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 $[0,1]$