I am confused on how to write proofs for geometric sums.
I think that using the well ordering principle to find the least n $\in$ $\mathbb{N}$ with $\alpha(n)$ $\ne$ $\beta(n)$ would be a good approach. I am not sure how to continue on from then. How would I write a proof for this example?
Suppose $\alpha$ $\ne$ $\beta$ $\in \{0, 2\}^\mathbb{N}$ Prove that $$\sum\limits_{k = 0}^\infty\frac{\alpha(k)}{3^k} \ne \sum\limits_{k = 0}^\infty\frac{\beta(k)}{3^k}. $$
Let $x$ be the first sum, and let $y$ be the second. Let $m$ be the smallest integer such that $\alpha_m \ne \beta_m$. (This is the start you suggested.)
Without loss of generality we may assume that $\alpha_m=0$ and $\beta_m=2$. Let $w=\sum_{i=0}^{m-1} \frac{\alpha_i}{3^i}=\sum_{i=0}^{m-1} \frac{\beta_i}{3^i}$. Then $$x\le w +\sum_{i=m+1}^\infty \frac{2}{3^i}=w+\frac{1}{3^m},$$ and $$y\ge w+\frac{2}{3^m},$$ and therefore $x\lt y$.