Using the floor function of $\frac{n}{k}$ leads the terms of the infinite series on the left to alternate sign every k terms. I tried several values of k and the following seems to be the case.
$$ \sum_{n = 0}^{\infty} \frac{\left(-1\right)^{\lfloor\,{n/k}\,\rfloor}}{x^n} = \frac{x\left(x^{k} - 1\right)}{\left(x - 1\right)\left(x^{k} + 1\right)}\,,\qquad \mbox{for}\ \left\vert x\right\vert \gt 1\ \mbox{and}\ k \in \mathbb{N}^{*} $$
The answer seems to relate to both the finite geometric series $\sum_{m = 1}^{k}x^{m} = x\left(x^{k} - 1\right)/\left(x - 1\right)$ and to $\sum_{n = 0}^{\infty}1/x^{n} = x/\left(x - 1\right)$ but I have no idea how trivial a result this is or how to prove it. (I found this while fooling around on wolfram and thought it looked interesting but I’m not a mathematician, it’s not homework)
Are there any other known series like this that alternate sign in interesting ways? For example using $(-1)^{(n+2\mod3)}$ for $++-++-++-$ etc..
Let $S$ denote the sum, then we can cancel out all but the first $k$ terms by adding $x^{-k}S$. $$S + x^{-k}S = \sum_{n=0}^{k-1} \frac{1}{x^n}$$ This is a geometric sum given by: $$\begin{align} \sum_{n=0}^{k-1} \frac{1}{x^n} &= \frac{ 1-x^{-k}}{1-x^{-1}} \\ &= \frac{x(1-x^{-k})}{x-1} \end{align}$$ Thus, $$(1 + x^{-k})S = \frac{x(1-x^{-k})}{x-1}$$ $$\Rightarrow S = \frac{x(1-x^{-k})}{(x-1)(1 + x^{-k})}$$ Multiplying the numerator and denominator by $x^{k}$ gives your expression: $$\Rightarrow S = \frac{x(x^{k}-1)}{(x-1)(x^{k}+1)}$$
For say $(-1)^{(n+2\bmod 3)}$, we would take the exact same approach.
$$S - x^{-3}S = 1 - x^{-1} + x^{-2}$$ $$\Rightarrow S = \frac{1 - x^{-1} + x^{-2}}{1-x^{-3}}$$
This is the same approach used to prove the sum of a geometric series and alternating series $(-1)^n$. We subtract/add some multiple of the summation to cancel out the tail of the series or at least make it more manageable.