# A better approximation of $({n \over e})^n$ than by $\Gamma(1+n-1/2)$ ? (Focus is “reverse” to the Stirling approximation)

In a formula in my self-study of a summation method based on the matrix of Eulerian numbers (which I thus call "Eulerian summation") I am considering terms like $$\Big({n \over e}\Big)^n \cdot {1 \over n!} \cdot {1 \over n+1} \tag 1$$ For the estimate of the growth there is the Stirling-approximation-formula for the factorial: $$m! \approx \sqrt{2 \pi m} \Big({m \over e}\Big)^m \tag 2$$

so that I arrive at the estimate

$$\Big({n \over e}\Big)^n {1 \over n!} \approx {n! \over \sqrt{2 \pi n} }\cdot {1 \over n!} = {1 \over \sqrt{2 \pi n} } \tag 3$$
(where I omit here and in the following the ${1 \over n+1}$-term)

But by inspection of some results with increasing n I found, that a much better approximation than(2) is $$\Big({n \over e}\Big)^n \approx \Gamma(1+n-1/2) \cdot {1 \over \sqrt{2 \pi }} \tag 4$$ From this, and from the coincidence that $\sqrt{\pi} = \Gamma(1/2)=\Gamma(1-1/2)$ one might then write $$\Big({n \over e}\Big)^n {1 \over n!} \approx {\Gamma(1+n-1/2) \over \Gamma(1-1/2) \cdot \Gamma(1+n) }\cdot {1 \over \sqrt{2}}\\ = \binom{n-1/2}{n}\cdot {1 \over \sqrt{2} } \tag 5$$
which is then a simple (generalized) binomial-expression.
By comparision , the series $$s = \sum_{n=0}^\infty \Big({n \over e}\Big)^n {1 \over n!} \cdot {1 \over n+1} \tag 6$$ should then be a convergent expression and nicely approximated by. $$t = {1\over \sqrt 2}\sum_{n=0}^\infty \binom{n-1/2}{n} \cdot {1 \over n+1} \sim s\tag 7$$

I thought, after that improvement the formula (4) might be even more improvable, so I looked first at Peter Luschny's "factorial" site where I find a related expression at the paragraph "A simple expression(recommended)" with the "LuschnyCF4" where however $m-1/2$ is replaced by $n$ and $n+1/2$ by $N$ and the approximation reads then as $$n! \approx \sqrt{2 \pi } \Big({P \over e}\Big)^N \\ \text{where } N=n+1/2 \tag 8$$ Here $P$ is a complicated scaling of $N$ so I cannot simply adapt that formula for an improved approximation. So my question is:

Q: How can I improve my (heuristical) approximation in the rhs in (4) (while keeping it focused at positive integers $n$ in the term $(n/e)^n$)?

(P.s.: I did not have a good idea for tagging, perhaps someone could improve that)

• I'm a little confused on what you are asking here. Are you looking for a proof of $(4)$? What do you mean by "while keeping it focused at positive integers $n$"? – EuYu Jul 4 '14 at 12:17
• Isn't simpler to just plugin the next terms of the Stirling approximation? – leonbloy Jul 4 '14 at 12:23
• @EuYu: hmm, in the Stirling-formula the $n$ is also under the square-root; in (4) it has moved mystically into the gamma-expression, so this is a variant of the Stirling-formula. Maybe I've not well expressed what I wanted to say. My problem is that I have the lhs in (4) and at integer n. Then I found that heuristically I can express that by Gamma( f(n)) where f(n) is some function of n (in this case f(n)=n+1/2) . After this ansatz, I can't see how to get an even better approximation by a variation of the expression on the rhs. Hope I could make it clearer? – Gottfried Helms Jul 4 '14 at 12:32
• @Leonbloy: please see my comment to EuYu – Gottfried Helms Jul 4 '14 at 12:34
• @GottfriedHelms I'm not sure if this helps, but Stirling's approximation doesn't just hold for integer $n$. It's in fact the first term of the asymptotic approximation for $\Gamma(n)$. If you substitute $n+\frac{1}{2}$ into Stirling's formula, you should be able to reduce it (asymptotically) to $(4)$. The fact that the $n$ has moved out of the square root is effectively because $\Gamma(n+\frac{1}{2})$ has introduced another $\sqrt{n}$ to the approximation. – EuYu Jul 4 '14 at 12:43

The asymptotic expansion for $\Gamma(z)$ is given by $$\Gamma(z) \sim \sqrt{\frac{2\pi}{z}}\left(\frac{z}{e}\right)^z\left(1 + \frac{1}{12z} + \cdots\right)$$ Therefore we have \begin{align}\Gamma\left(n+\frac{1}{2}\right) &\sim \sqrt{\frac{2\pi}{\left(n+\frac{1}{2}\right)}}\left(\frac{n+\frac{1}{2}}{e}\right)^{n+\frac{1}{2}} \\&= \sqrt{2\pi}\frac{\left(n+\frac{1}{2}\right)^n}{e^{n+\frac{1}{2}}} \\&= \sqrt{2\pi}\left(\frac{n}{e}\right)^n\frac{\left(1+\frac{1}{2n}\right)^n}{\sqrt{e}}\end{align}
Now since we have $$\lim_{n\rightarrow \infty}\left(1 + \frac{x}{n}\right)^n = e^x$$ It follows that $$\left(1+\frac{1}{2n}\right)^n \sim \sqrt{e}$$ And therefore we recover your approximation $(4)$ with the above substitution: $$\Gamma\left(n+\frac{1}{2}\right) \sim \sqrt{2\pi}\left(\frac{n}{e}\right)^n\frac{\left(1+\frac{1}{2n}\right)^n}{\sqrt{e}} \sim \sqrt{2\pi}\left(\frac{n}{e}\right)^n$$ Of course the asymptotics can be kept more accurate without making the latter approximation, but that comes at the expensive of a more complex expression. Also, note that you can obtain increasingly accurate asymptotics by taking more and more terms from the asymptotic series.