A new infinite series for the minimum of the Gamma function? I continue my work on the minimum of the Gamma's function and recently I found a slightly different way via an infinite series as I have the beginning :
Let $0<x$ then define :
$$f(x)=x!,h(x)=f'(x)$$
Then we have :
$$h(k)=0$$
And :
$$k\simeq \frac{1}{2}\left(\pi-e+\frac{1}{2}\right)-\frac{1}{28}\left(\pi-e\right)^{\pi e}-\frac{10}{37}\left(\pi-e\right)^{2\pi e}-\frac{10000}{3675}\left(\pi-e\right)^{3\pi e}-\frac{1000}{499}\left(\pi-e\right)^{4\pi e}$$
I would like to find a infinite series like :
$$k=\frac{1}{2}\left(\pi-e+\frac{1}{2}\right)-\sum_{k=1}^{\infty}a_k\left(\pi-e\right)^{k\pi e}$$
Where $a_k>0$
I can progress numericaly not theoreticaly .
Edit following Tyma Gaidash's comment :
We have (if there is no mistake) 36 decimals right for the minimum value of the gamma function taking for $k$ :
$$k\simeq \frac{1}{2}\left(\pi-e+\frac{1}{2}\right)-\frac{1}{28}\left(\pi-e\right)^{\pi e}-\frac{10}{37}\left(\pi-e\right)^{2\pi e}-\frac{10000}{3675}\left(\pi-e\right)^{3\pi e}-\frac{1000}{499}\left(\pi-e\right)^{4\pi e}-\frac{736}{100}\left(\pi-e\right)^{5\pi e}-\frac{5}{10}\left(\pi-e\right)^{6\pi e}$$
How to find the sequence $a_k$ ?
 A: Assume arbitrary constants $x_0,\xi\in\textbf{R}$. Then exists constants $a_0,a_1,a_2,\ldots$ and analytic function $f(x)$ such that (take for example $a_k=\left(\frac{x_0-1}{\xi x_0}\right)^k$):
$$
x_0=\sum^{\infty}_{k=0}a_k\xi^k\tag 1
$$
and
$$
\frac{f^{(k)}(\xi)}{k!}=a_k\tag 2
$$
and
$$
f(2\xi)=x_0.\tag 3
$$
We expand $f$ into power series
$$
f(x)=\sum^{\infty}_{k=0}\frac{f^{(k)}(\xi)}{k!}(x-\xi)^k\Rightarrow
f(x+\xi)=\sum^{\infty}_{k=0}\frac{f^{(k)}(\xi)}{k!}x^k\Rightarrow
f(2\xi)=\sum^{\infty}_{k=0}\frac{f^{(k)}(\xi)}{k!}\xi^k\Rightarrow
$$
$$
x_0=\sum^{\infty}_{k=0}a_k\xi^k
$$
Set now $\xi=\frac{1}{2}(\pi-e)^{\pi e}$. Then $f\left((\pi-e)^{\pi e}\right)=x_0$.
However, I want not make fun of the problem. To find such $a_k$ is a very great achievement. For example in [B] page 194 Entry 15, Ramanujan give us a way to construct such series:
Set
$$
\log a=\psi(x+1),
$$
then
$$
\left(\frac{x+\frac{1}{2}}{a}\right)^{4n}\approx 1-\frac{n}{6a^2}+\frac{10n^2+11n}{720a^4}-\frac{70n^3+231 n^2+891 n}{90720a^6}+\ldots\tag 4
$$
When $x=x_0$ i.e. as in your problem, then equivalently $x_0$ is root of $\psi(x+1)=0$. Hence $a=1$ and we get approximations of $x_0$ using (4).
References
[B]: Bruce C. Berndt. "Ramanujan's Notebooks I". Springer-Verlag.
New York, Berlin, Heidelberg, Tokyo, 1985.
