During a homework assignment, I ran into this series: $$S(x) = \sum_{n=0}^\infty \frac{n x^n}{1 - \left( \frac{x}{1+x} \right)^n}$$ Graphing $S(x)$ shows that it converges for some values of $x$, mainly near $x=0$ and near $x=-1$. I was wondering if $S$ has a closed form. Despite my best efforts, I was not able to find one online or through my own analysis.
It should be noted that the summand is ill-defined for $n=0$. It does, however, possess a limit. In particular: $$\lim_{n \to 0} \! \left ( \frac{n x^n}{1 - \left( \frac{x}{1+x} \right)^n} \right ) = \frac{1}{\ln \! \left( 1+\frac{1}{x} \right)}$$ This limit should be taken as the $n=0$ term.
This is not an answer.
The continuous case is interesting $$ T(x) = \int_{1}^\infty \frac{n\, x^n}{1 - y^n}\,dn=\int_{1}^\infty\frac{n\, e^{a n}}{1-e^{b n}}\,dn$$ with $$ \qquad 0 < x < 1\qquad \qquad y=\frac x{1+x}\qquad \qquad a=\log(x)\qquad \qquad b=\log(y)$$
$$\large\color{blue}{T(x)=\frac{e^a}{a\, b^2} \left(a\, \Phi \left(e^b,2,\frac{a}{b}\right)-b^2 \,\, _2F_1\left(1,\frac{a}{b};\frac{a+b}{b};e^b\right)\right)}$$
If we compare with $$ S(x) = \sum_{n=1}^\infty \frac{n\, x^n}{1 - y^n}$$ (for $0.05 \leq x \leq 0.95$), we have a correlation coefficient $R=0.999938$.