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)
 A: 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.
A: Legendre's duplication formula implies that
$$\tag{1}
\log \Gamma \!\left( {z + \frac{1}{2}} \right) = \log \Gamma (2z) - \log \Gamma (z) - (2z - 1)\log 2 + \frac{1}{2}\log (2\pi )
$$
for $|\arg z|<\pi$. The asymptotic expansion of the logarithm of the gamma function reads
$$\tag{2}
\log \Gamma (z) \sim \left( {z + \frac{1}{2}} \right)\log z - z + \frac{1}{2}\log (2\pi ) + \frac{1}{{12z}} - \frac{1}{{360z^3 }} +  \cdots 
$$
as $z\to \infty$ in the sector $|\arg z|\leq \pi-\delta$ ($<\pi)$. Employing $(2)$ in $(1)$ gives
$$\tag{3}
\log \Gamma \!\left( {z + \frac{1}{2}} \right) \sim z\log z - z + \frac{1}{2}\log (2\pi ) - \frac{1}{{24z}} + \frac{7}{{2880z^3 }} -  \cdots 
$$
as $z\to \infty$ in the sector $|\arg z|\leq \pi-\delta$ ($<\pi)$. By taking the exponential of each side and expanding the exponential of the asymptotic series, we deduce
$$
\Gamma\! \left( {z + \frac{1}{2}} \right) \sim \left( {\frac{z}{e}} \right)^z \sqrt {2\pi } \left( {1 - \frac{1}{{24z}} + \frac{1}{{1152z^2 }} + \frac{{1003}}{{414720z^3 }} - \frac{{4027}}{{39813120z^4 }} -  \cdots } \right)
$$
as $z\to \infty$ in the sector $|\arg z|\leq \pi-\delta$ ($<\pi)$. If you solve for $z\log z-z$ in $(3)$ and do the exponentiation afterwards, you will find
$$
\left( {\frac{z}{e}} \right)^z \sqrt {2\pi } \sim  \Gamma\! \left( {z + \frac{1}{2}} \right)\left( {1 + \frac{1}{{24z}} + \frac{1}{{1152z^2 }} - \frac{{1003}}{{414720z^3 }} - \frac{{4027}}{{39813120z^4 }} +  \cdots } \right)
$$
as $z\to \infty$ in the sector $|\arg z|\leq \pi-\delta$ ($<\pi)$. The reason for having the same coefficients (apart from the signs) is coming from the fact that $(3)$ is a series in odd negative powers of $z$.
It is possible to expand in gamma functions with decreasing arguments as well:
$$
\left( {\frac{z}{e}} \right)^z \sqrt {2\pi }  \sim \Gamma \left( {z + \frac{1}{2}} \right) + \frac{1}{{24}}\Gamma \left( {z - \frac{1}{2}} \right) - \frac{{23}}{{1152}}\Gamma \left( {z - \frac{3}{2}} \right) +  \cdots .
$$
This expansion has been used recently in https://arxiv.org/abs/1907.03543
