Naively estimating the factorial

Question

A naive way to estimate the factorial is $n! \geq (a+1) (a+2) \dots n \geq a^{n-a}$ for any $a$. For example, it gives $n! \geq (n/2)^{n/2}$ and slightly better $n! \geq (n/3)^{2n/3}$. I am interested in how strong this naive estimation it can be. For which $a$ we get the best estimate?

If my calculations are correct, this happens when $a = \frac{n}{W(e n)}$ where $W$ is Lambert W. Can the expression $\left(\frac{n}{W(e n)}\right)^{n-\frac{n}{W(e n)}}$ be simplified, or sensibly estimated by elementary functions?

Answer 1

Using Stirling's approximation we have $n! \approx (\frac ne)^n$. If we take your $a$ to be a fixed fraction of $n$, say $\frac nb$, your gives $(\frac nb)^{n(1-\frac 1b)}$. Stirling divided by yours is then $ \dfrac {(\frac ne)^n}{(\frac nb)^{n(1-\frac 1b)}}=\dfrac {b^{n(1-\frac 1b)}}{n^{\frac nb}}\lt \left(\dfrac b{n^\frac 1b}\right)^n$ which goes to zero with large $n$ for any $b$, quickly when $n \gt b^b$

Answer 2

I will only try to reproduce your computation. We are trying to maximize $f(a):=a^{n-a}$ for $a\in[0,n-1]$. The derivative is $$f'(a)=a^{n-a}[-\ln(a)+\frac{n-a}{a}].$$

We equate to zero

$$\ln(a)=\frac{n-a}{a}=\frac{n}{a}-1,$$

i.e. $$\ln(ea)=\frac{n}{a}.$$

Taking exponential $ea=e^{n/a}$, i.e. $\frac{n}{a}e^{\frac{n}{a}}=en$. From this $n/a=W(en)$ and $a=n/W(en)$. It doesn't mean much by I got the same.

Now, to estimate your expression involving $W$ one can use asymptotic expansions of $W$.

We have that $$\left(\frac{W(z)}{z}\right)^r=\sum_{n=0}^\infty \frac{r(k+r)^{k-1}}{k!}\ (-z)^k.$$

Let us put this in $H(n):=e^{(n-n/W(en))\ln(n/W(en))}$ for $r=-1$.

No, this one is not good. It holds for $|z|<e^{-1}$. We need asymptotics to infinity. Here they have some (Equation 4.20). Then those are the ones we need to put in the formula for $H$.