If $z$ is a positive irrational number and I have the series $$\sum_{n=0}^{k} z^{n} $$ are there conditions which this will or will not sum to a positive rational?
I am having trouble seeing the conditions when the sum could be irrational and when rational.
For instance let $y \in \mathbb Q $ if I plug in the following equation into wolfram alpha $$1+z+z^2+z^3+z^4=2y-1$$ and I tell it that y is rational it answers that $z \in \mathbb Q$ but I do not see why $z$ could not be irrational?
We have
$$\sum_{k=0}^{n} z^{k}=\frac{z^{n+1}-1}{z-1}$$
This can indeed be rational for non-rational (although complex) $z$. For example set $z=\exp\left(i\frac{2\pi}{n}\right)$. Then
$$\frac{z^{n+1}-1}{z-1}=\frac{\exp\left(i\frac{2\pi}{n}(n+1)\right)-1}{\exp\left(i\frac{2\pi}{n}\right)-1}=\frac{1\cdot\exp\left(i\frac{2\pi}{n}\right)-1}{\exp\left(i\frac{2\pi}{n}\right)-1}=1$$
In general, we have that for rational $r$ the following holds
$$\frac{z^{n+1}-1}{z-1}=r$$
$$\Leftrightarrow p(z)=z^{n+1}-rz-1+r=0$$
This implies that the sum is rational if and only if $z$ is a root of the this polynomial. Can we describe this set based only on the parameter $r$? This is a much harder question. Obviously, we can just state
$$S=\{z:p(z)=0\text{ for some rational }r\}$$
and be done. However, this is a slightly unsatisfying answer. A better answer might be solving this polynomial for an arbitrary $n$ and then presenting the solutions. Is this always possible? I'm not sure, obviously it can be done for $n\leq 4$ and mathematica spits out a complicated answer for $n=5$ but I don't know enough to determine if $p(z)$ is solvable over radicals for any $n$ and $r$.