Suppose $\alpha$ $\ne$ $\beta$ $\in \{0, 2\}^\mathbb{N}$ Prove that $$\sum\limits_{k = 0}^\infty\frac{\alpha(k)}{3^k} \ne \sum\limits_{k = 0}^\infty\frac{\beta(k)}{3^k}. $$
This is the written proof I found online:
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 $x\lt y$.
I am having trouble understanding the logic from the "Then x $x\le w +\sum_{i=m+1}^\infty \frac{2}{3^i}=w+\frac{1}{3^m},$" How does the logic work for this proof?
The sum $w$ covers the first $m$ terms, where the two sums agree. What is left over is the sum from the $m$th term on. We have $\alpha_m = 0$ so we can write $$x = w + \sum_{i=m+1}^\infty \frac{\alpha(i)}{3^i}$$ But of course since $\alpha(i)$ is either $0$ or $2$, $\alpha(i) \leq 2$
$$x \leq w + \sum_{i=m+1}^\infty \frac{2}{3^i}$$ The infinite part of the right hand side is a geometric series.
On the other hand, $$y = w + \frac{\beta(m)}{3^m}+ \sum_{i=m+1}^\infty \frac{\beta(i)}{3^i}$$ By assumption $\beta(m) = 2$ and the worst case scenario is that all future $\beta$ are $0$, so $$y \geq w + \frac{2}{3^m} + 0$$
Does this address your question?