The following power series:
$$f(x)=\sum _{n=1}^{\infty } \frac{x n^{k n} (-x)^n (k!)^n}{(n+1) (k n)!}+x+1$$
is equal to x/LambertW(x) when expanded at $x=0$ and when k=1. As pointed out by Andre LeClaire, the interesting x values appear to be:
$$x=\frac{1-\frac{11}{8}}{\exp (1)}$$ and: $$x=\frac{2-\frac{11}{8}}{\exp (1)}$$
robjohn (in the chat room) gave this approximate radius of convergence:
$$r=\frac{1}{\sqrt{2 \pi k}}$$
Solving for $k$ in: $$-\frac{1-\frac{11}{8}}{\exp (1)}=\frac{1}{\sqrt{2 \pi k}}$$
and:
$$\frac{2-\frac{11}{8}}{\exp (1)}=\frac{1}{\sqrt{2 \pi k}}$$
we get in the first case:
$$k = \frac{32 e^2}{9 \pi }$$
and in the second case:
$$k = \frac{32 e^2}{25 \pi }$$
But if the radius of convergence is only approximately:
$$r=\frac{1}{\sqrt{2 \pi k}}$$
then these values of $k$ above are not the exact values I am interested in, because what I would like to do is to maximize $k$ in $f(x)$ while the value of the $k$-dependent radius of convergence needs to suit the values of $x$ above. After $k$ has been maximized I would like to see what happens if $11/8$ in the $x$ is changed to some other number. The goal is to find an approximation for the first and second non-trivial zeta zeros.
So my question is, what is the exact radius of convergence for the power series $f(x)$?
Use the root test: $$ \Bigl(\frac{n^{kn}(k!)^n}{(n+1)(k\,n)!}\Bigr)^{1/n}=\frac{n^k\,k!}{(n+1)^{1/n}((k\,n)!)^{1/n}}. $$ By Stirling's formula $$ (k\,n)!\sim\Bigl(\frac{k\,n}{e}\Bigr)^{kn}\sqrt{2\,\pi\,k\,n} $$ so that $$ ((k\,n)!)^{1/n}\sim\Bigl(\frac{k}{e}\Bigr)^{k}n^k(2\,\pi\,k\,n)^{1/2n}. $$ It follows that $$ \lim_{n\to\infty}\Bigl(\frac{n^{kn}(k!)^n}{(n+1)(k\,n)!}\Bigr)^{1/n}=\frac{k!}{(k/e)^k} $$ and $$ r=\frac{1}{k!}\Bigl(\frac{k}{e}\Bigr)^{k}. $$ Sanity check: if $k$ is large, again by Stirling's formula $$ r\sim\frac{1}{\sqrt{2\,\pi\,k}}. $$