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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?

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thestar
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    Have you applied the geometric series summation formula to this problem? – Lee Mosher Aug 25 '21 at 16:11
  • @BarryCipra thank you – thestar Aug 25 '21 at 16:44
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    It would help if you explained in some detail exactly what you are plugging into wolfram alpha, because it seems highly unlikely that WA would tell you that rational values of $y$ in your example displayed equation give rational values of $z$. That equation is a quartic polynomial in $z$, which in general does not have rational roots. – Barry Cipra Aug 25 '21 at 16:51
  • @BarryCipra this is what I entered "1+x+x^2+x^3+x^4=2y-1, y is rational" – thestar Aug 25 '21 at 16:52
  • And what exactly was WA result? When I tried it just now, it told me "$y={1\over2}(x^4+x^3+x^2+x+1)$ and $x\in\mathbb{Q}$." I think all alphie is doing is telling you how to ensure that $1+x+x^2+x^3+x^4=2y-1$ and $y$ is rational, not how to get all such $y$'s. – Barry Cipra Aug 25 '21 at 16:59
  • Ahh so x could be irrational – thestar Aug 25 '21 at 17:00
  • @thestar, yes, exactly. You were just misinterpreting WA's (slighly cryptic) answer. – Barry Cipra Aug 25 '21 at 17:04

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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 $r$.

QC_QAOA
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    I can tell you the $r$ that I am looking at specifically are of the form $2x^2-1$, for positive integer x @QC_QAOA – thestar Aug 25 '21 at 17:19
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    Point still stands, there are irrational, positive $z$ but you might be able to do better than numeric approximations. – QC_QAOA Aug 25 '21 at 17:31