How to understand base-n expansions of numbers less than 1?

Solution 1:

As @Nikolaj said the ambiquity is the same as in base $10$.

About why $x = \sum_{k - 1}^{\infty} \ a_k \ 3^{-k} \ \mid a_k \in \{0, 1, 2\}$:
How we represent a decimal number in base $10$ as a sum? Let $x = 0.15$, we can say:
$x = 0.1 + 0.05 = 1 \cdot 10^{-1} + 5 \cdot 10^{-2} + 0 \cdot 10^{-3} + \cdots \\ = \sum_{k - 1}^{\infty} \ a_k \ 10^{-k} \ \mid a = \{1, 5, 0, 0, \cdots\}$

The same happens in base $3$.

A example of the ambiguity in the sum:
$\sum_{k}^{\infty} \ a_k \ 10^{-k} \ \mid a = \{1, 0, 0, \cdots\} = \sum_{k - 1}^{\infty} \ a_k \ 10^{-k} \ \mid a = \{2, 2, 2, \cdots\} \\ \Rightarrow 1 = 0.\bar{2}$
This is true for any base $n$: $x = (x - 1).\overline{(n - 1)}$