Maxima of $f(x)/e^x$ where $f(x)$ is an approximation of $e^x$ using Stirling's Let $$f(x)=1+\sum_{n=1}^\infty\frac{x^n}{\sqrt{2\pi n}(n/e)^n}\tag1$$
and let $$g(x)=\frac{f(x)}{e^x}\tag2$$
If we plot $g(x)$ we get a graph that looks like this:

Clearly there is a maximum at around $x_0=1.5$, investigating further we find an approximation to this $x_0$ is $$x_0\approx 1.501418665538103821742229435035476066021$$
I did not manage to find a closed form using WolframAlpha or ISC.
Questions:


*

*Has a crude approximation to $e^x$ like this ever been studied? Specifically one using stirlings approximation.

*Although I doubt it, is there any observation that can be mind that might hint at the actual value of $x_0$?

*Is there a good way of numerically approximating $x_0$?

 A: $\displaystyle n!=\sqrt{2\pi n}(\frac{n}{e})^n A(n)$ $\quad$  with $\quad$ $\displaystyle A(n):=\prod\limits_{k=0}^\infty ((1+\frac{1}{k+n})^{k+n+\frac{1}{2}}/e)$  
$\displaystyle f(x):=1+\sum\limits_{n=1}^\infty \frac{x^n}{n!}A(n)$ $\quad$ with $\quad$ $\displaystyle A(1)=\frac{e}{\sqrt{2\pi}}$ $\quad$ and $\quad$ 
$\displaystyle \frac{A(n)}{A(n+1)}=(1+\frac{1}{n})^{n+\frac{1}{2}}/e$
Maximum for $\displaystyle e^{-x}f(x)$: $\quad$ $\displaystyle (f(x)e^{-x})´=f´(x)e^{-x}-f(x)e^{-x}:=0$ 
=> $\quad$ $\displaystyle f´(x)=f(x)$ $\quad$ which means $\quad$ $\displaystyle 1+\sum\limits_{n=1}^\infty \frac{x^n}{n!}A(n)= A(1)+\sum\limits_{n=1}^\infty \frac{x^n}{n!}A(n+1)$.
An iteration to get the value $x$ is e.g. 
$$\displaystyle x_{k+1}=\frac{A(1)-1}{\sum\limits_{n=1}^\infty\dfrac{x_k^{n-1}}{n!}(A(n)-A(n+1))}$$ with $\quad$ $x_0:=1,5$. 
$\displaystyle A(n)-A(n+1)$ converges quickly, therefore there are not many members of the series $\sum\limits_{n=1}^\infty\dfrac{x^{n-1}}{n!}(A(n)-A(n+1))$ for the iteration necessary to get a good result for $x$. 
