What can we say about $\sum_{m=1}^{\infty} r^{m^s}= ?$ for $|r|<1.$ We know the geometric series:
$$\sum_{m=1}^{\infty} r^{m}= \frac{r}{1-r}$$ for $|r|<1.$
(Please correct me if I've made mistake.)

My question is: For fix $s>0,$
  What can we say about $$\sum_{m=1}^{\infty} r^{m^s}= ?$$ for $|r|<1.$

 A: It is hopeless to expect a closed form when $s \neq 1$. Still, we may extract some properties of the sum. For instance, if $s > 0$ and $r \in (0, 1)$, then by writing $\epsilon = -\log r$, we have
\begin{align*}
\sum_{m=1}^{\infty} r^{m^s}
= \sum_{m=1}^{\infty} e^{-\epsilon m^s}
= \sum_{m=1}^{\infty} \int_{\epsilon m^s}^{\infty} e^{-x} \, \mathrm{d}x
= \int_{0}^{\infty} \left( \sum_{m=1}^{\infty} \mathbf{1}_{\{x \geq \epsilon m^s\}} \right) e^{-x} \, \mathrm{d}x.
\end{align*}
Now by writing
\begin{align*}
\sum_{m=1}^{\infty} \mathbf{1}_{\{x \geq \epsilon m^s\}}
= \sum_{m=1}^{\infty} \mathbf{1}_{\{m \leq (x/\epsilon)^{1/s} \}}
= \sum_{m=1}^{\lfloor (x/\epsilon)^{1/s} \rfloor} 1
= \lfloor (x/\epsilon)^{1/s} \rfloor
\end{align*}
and plugging this back, we get
\begin{align*}
\sum_{m=1}^{\infty} r^{m^s}
= \int_{0}^{\infty} \lfloor(x/\epsilon)^{1/s}\rfloor e^{-x} \, \mathrm{d}x.
\end{align*}
(If you are familiar with Stieltjes integral, you can obtain this integral representation quite easily as follows:
\begin{align*}
\sum_{m=1}^{\infty} e^{-\epsilon m^s}
= \int_{(0,\infty)} e^{-x} \, \mathrm{d}\lfloor (x/\epsilon)^{1/s} \rfloor
=  \int_{0}^{\infty} \lfloor(x/\epsilon)^{1/s}\rfloor e^{-x} \, \mathrm{d}x,
\end{align*}
where the second step follows from integration by parts.) Without the flooring function, we would get
\begin{align*}
\int_{0}^{\infty} (x/\epsilon)^{1/s} e^{-x} \, \mathrm{d}x
= \frac{\Gamma(1+\frac{1}{s})}{\epsilon^{1/s}}
\end{align*}
by using the gamma function. Instead, using the inequality $x-1 \leq \lfloor x \rfloor \leq x$, we obtain
\begin{align*}
\frac{\Gamma(1+\frac{1}{s})}{\epsilon^{1/s}} - 1
\leq \sum_{m=1}^{\infty} e^{-\epsilon m^s}
\leq \frac{\Gamma(1+\frac{1}{s})}{\epsilon^{1/s}}
\end{align*}
for $r \in (0, 1)$ and $s > 0$. From this we may extract some asymptotic behavior of the sum. Similar argument shows that, if $s$ is a positive integer and $r \in (0, 1)$, then
\begin{align*}
\sum_{m=1}^{\infty} (-r)^{m^s}
= - \int_{0}^{\infty} (\lfloor(x/\log(1/r))^{1/s}\rfloor \text{ mod } 2) e^{-x} \, \mathrm{d}x
\xrightarrow[\text{as } r\uparrow 1]{} -\frac{1}{2}.
\end{align*}
