# Convergence of Geometric Series for irrational numbers

I was looking at the convergence of geometric series for irrational values for $a$ and $r$ such that I let $r=\frac 1a$. I noticed some (most likely trivial) pattern and I wonder if there is some way that I can generalize the results that I get.

I started with the obvious $a=\sqrt 2$ and then let $r=\frac 1{\sqrt 2}$ and get

$$S_2 \;=\; \sum_{n-1}^{\infty}ar^{n-1}=2+2\sqrt 2$$

I want to show that $S_2$ is irrational. I know how to show that $\sqrt2 \not \in \Bbb Q$ and from there it seems clear to me that $2+2\sqrt2$ has to be irrational since adding $2$ and multiplying by $2$ does not change the fact that the decimal expansion of $\sqrt 2$ is infinite.

That seems like a rather weak argument to me though, how can I express this formally?

Then, I went on to try looking at the geometric series for some of the next obvious irrationals for $a$ and $r$ and got the following results:

$$a=\sqrt 3, \, r=\frac 1{\sqrt3} : \; S_3 \;=\; \sum_{n-1}^{\infty}ar^{n-1}= \frac 12 (3 + 3 \sqrt 3)$$

$$a=\sqrt 5, \, r=\frac 1{\sqrt5} : \; S_5 \;=\; \sum_{n-1}^{\infty}ar^{n-1}=\frac 14 (5 + 5 \sqrt 5)$$

$$a=\sqrt 7, \, r=\frac 1{\sqrt7} : \; S_7 \;=\; \sum_{n-1}^{\infty}ar^{n-1}=\frac 16 (7 + 7 \sqrt 7)$$

$$a=\sqrt 8, \, r=\frac 1{\sqrt8} : \; S_8 \;=\; \sum_{n-1}^{\infty}ar^{n-1}=\frac 17 (8 + 8 \sqrt 8)$$

It looks like when $a \not \in \Bbb Q \, \wedge \,x\in \Bbb Z^+ : \; a = \sqrt x \; \wedge\; r=\frac 1a$ then the series $$S_x \;=\; \sum_{n-1}^{\infty}ar^{n-1}=\frac 1{a^2-1} (a^2 + a^3) \,:\, S_x \not \in \Bbb Q$$

I thought of using induction to prove this but since this is a statement about irrational numbers I does not make sense. Furthermore, induction on $x \in \Bbb Z^+$ for $S_x$ would inevitably pass trough rational values for $a$. (e.g. It fails at $x=4$ and $S_4$ converges to $4$.)

So, I am looking to show that these series I computed all converge to irrational values for $S_x$.

Then, from there I would like to know whether it can be shown that in general, the following is true, $$\left(\sum_{n-1}^{\infty}ar^{n-1}=S_x \,:\, S_x \not \in \Bbb Q \right)\;\left(\forall\, a \not \in \Bbb Q \, \wedge \,x\in \Bbb Z^+ : \; a = \sqrt x \; \wedge \; r=\frac 1a\right)\,.$$

For the first part, we know that $\Bbb Q$ is a field, which is a fancy way of saying it is closed under addition and multiplication, among other things. Thus if $2+2\sqrt{2}\in \Bbb Q$, then so would $\frac12(2+2\sqrt{2}-2)=\sqrt{2}$, a contradiction.
To see that your formula holds, $$\sum_{n=0}^{\infty}\sqrt{x}x^{-\frac{n}{2}} = \sqrt{x}\frac{1}{1-\frac{1}{\sqrt{x}}}=\frac{x}{\sqrt{x}-1}=\frac{1}{x-1}(x\sqrt{x}+x)$$ If this was in the rational numbers, then we could multiply by $x-1$, subtract by $x$, and then divide by $x$, so the sum is rational only when $a$ is rational, which only holds when $x$ is a perfect square.