Does $\sum^\infty_{s=1}\delta^{s-1}\frac{1}{1 + cs}$ have a closed-form expression? 
I'm wondering whether
$$
\sum^\infty_{s=1}\delta^{s-1}\frac{1}{1 + c \cdot s},
$$
with $\delta \in (0,1)$ and $c\ge 0$ has a closed-form expression.

While I don't know of a general closed-form I can think if a special case:
Setting $c=0$ we obtain
$$
\sum^\infty_{s=1}\delta^{s-1}=\frac{1}{1-\delta},
$$
which is the geometric series. As for the more general case $c\neq 0$ I'm not sure.

For $|x^c|< 1$, consider the geometric series $\sum\limits_{n=0}^{\infty}x^{nc}$. Then we know
$$\frac{1}{1-x^c}=\sum\limits_{n=0}^{\infty}x^{nc}.$$
So
\begin{align*}
\frac{x^c}{1-x^c}& =\sum_{n=0}^{\infty}x^{nc+c}\\
\int\frac{x^c}{1-x^c} \, dx& =\sum_{n=0}^{\infty}\frac{x^{nc+c+1}}{1+c(n+1)}\\
\int\frac{x^c}{1-x^c} \, dx& =x^{c+1}\sum_{n=0}^{\infty}\frac{x^{nc}}{1+c(n+1)}.
\end{align*}
Now let $\delta=x^c$ (with $c \neq 0$), so we have
\begin{align*}
\frac{1}{c}\int\frac{\delta^{1/c}}{1-\delta} \, d\delta& =\delta^{\frac{c+1}{c}}\sum_{n=0}^{\infty}\frac{\delta^{n}}{1+c(n+1)}\\
\color{red}{\frac{1}{c\delta^{\frac{c+1}{c}}}\int\frac{\delta^{1/c}}{1-\delta} \, d\delta}& =\sum_{n=0}^{\infty}\frac{\delta^{n}}{1+c(n+1)}.
\end{align*}
So if we can solve the integral then we can get some closed form. However solving this integral may not be that straightforward.
