TITLE: A conjecture formula: $\sum\limits_{n=1}^\infty \frac{\binom{mn}{n}}{n}\left(\frac{(m-1)^{m-1}}{m^m} \right)^n=m\log\left(\frac{m}{m-1}\right)$
QUESTION [12 upvotes]: By the help of Mathematica numeral calculations, I find the following formula holds
$$\sum\limits_{n=1}^\infty \frac{\binom{mn}{n}}{n}\left(\frac{(m-1)^{m-1}}{m^m} \right)^n=m\log\left(\frac{m}{m-1}\right)\quad ?$$
$m>1$ is a positive integer. But I can't prove it.

REPLY [7 votes]: Let $z_m=(m-1)^{m-1}/m^m$. From this answer, we have
\begin{align}
F_m(z)&:=\sum_{n=0}^{\infty}\binom{mn}{n}\frac{z^n}{(m-1)n+1}=1+z\big(F_m(z)\big)^m,
\\ G_m(z)&:=\sum_{n=0}^{\infty}\binom{mn}{n}z^n=\frac{F_m(z)}{m-(m-1)F_m(z)}.
\end{align}
Now $F_m(0)=1$ and $\color{blue}{F_m(z_m)=m/(m-1)}$ (yeah!), thus
$$\sum_{n=1}^{\infty}\binom{mn}{n}\frac{(z_m)^n}{n}=\int_{0}^{z_m}\frac{G_m(z)-1}{z}\,dz,$$
and the substitution $w=F_m(z)$ (i.e. $z=(w-1)/w^m$) collapses it to
$$\sum_{n=1}^{\infty}\binom{mn}{n}\frac{(z_m)^n}{n}=\color{blue}{m\int_1^{m/(m-1)}\frac{dw}{w}}=m\ln\frac{m}{m-1}.$$
[As a by-product, we get $\displaystyle\sum_{n=1}^{\infty}\binom{mn}{n}\frac{z^n}{n}=m\ln F_m(z)$.]