Base-$b$ expansions: for an integer $b \ge 2$ every $x \in [0,1)$ is the sum of $\sum_{j \ge 0} d_j / b^{\,j+1}$ for digits $d_j < b$, and the digit sequence is unique among those that are not eventually constantly $b-1$

$\beta = \iota(b) \ge \iota(2) = 1 + 1 > 1 > 0$, and an induction gives $\beta^{\,n} > 0$ for every $n$; also $0 < 1/\beta < 1$.

For uniqueness, let $(c_j)$ and $(c'_j)$ be non-terminal digit sequences whose series have the same sum, and suppose they are not equal. By [L4] let $n$ be least with $c_n \ne c'_n$; interchanging the two sequences if necessary, assume $c_n < c'_n$, so $\iota(c'_n) - \iota(c_n) \ge 1$.

The digit of a residue. For every $r \in [0,1)$ there is exactly one natural $d < b$ with $\iota(d)/\beta \le r < \iota(d+1)/\beta$. For existence, the set $D := \{\, d \in \mathbb{N} : d \le b \text{ and } r < \iota(d)/\beta \,\}$ contains $b$, since $\iota(b)/\beta = 1 > r$; let $m := \min D$, which is not $0$ because $\iota(0)/\beta = 0 \le r$; put $d := m - 1$, so $d < b$, and minimality of $m$ says $d \notin D$, that is $\iota(d)/\beta \le r$, while $m \in D$ says $r < \iota(m)/\beta = \iota(d+1)/\beta$. For uniqueness, if $d < d'$ both worked then $r < \iota(d+1)/\beta \le \iota(d')/\beta \le r$, which the order forbids. Write $d(r)$ for this digit.

Both series converge, by hypothesis, so by linearity $\sum_j \bigl(\iota(c'_j) - \iota(c_j)\bigr)/\beta^{\,j+1}$ converges with sum $0$; its first $n$ terms vanish, so by [L9] the tail from $n$ also has sum $0$, that is $0 = \bigl(\iota(c'_n)-\iota(c_n)\bigr)/\beta^{\,n+1} + \sum_{j > n}\bigl(\iota(c'_j)-\iota(c_j)\bigr)/\beta^{\,j+1}$.

The residue map. For $r \in [0,1)$ put $f(r) := \beta r - \iota(d(r))$. Multiplying $\iota(d(r))/\beta \le r < \iota(d(r)+1)/\beta = (\iota(d(r)) + 1)/\beta$ by $\beta > 0$ gives $\iota(d(r)) \le \beta r < \iota(d(r)) + 1$, so $0 \le f(r) < 1$; thus $f$ is a function from $[0,1)$ to $[0,1)$.

Every difference satisfies $\iota(c'_j) - \iota(c_j) \ge -(\beta - 1)$, the digits lying in $\{0,\dots,b-1\}$, and $\sum_{j>n}(\beta-1)/\beta^{\,j+1} = \frac{\beta-1}{\beta^{\,n+2}}\cdot\frac{1}{1-1/\beta} = \frac{1}{\beta^{\,n+1}}$ by the geometric series; hence the tail in step 2.2 is at least $-1/\beta^{\,n+1}$.

By the recursion theorem applied to the set $[0,1)$, the element $x$ and the function $f$, there is a unique sequence $(r_n)$ in $[0,1)$ with $r_0 = x$ and $r_{n+1} = f(r_n)$; put $d_j := d(r_j)$, a natural number $< b$, so $(d_j)$ is a digit sequence in base $b$.

Write $A$ for the first summand and $B$ for the tail in step 2.2, so $A + B = 0$ while $A \ge 1/\beta^{\,n+1}$ by step 1.2 and $B \ge -1/\beta^{\,n+1}$ by step 3.2; since the two lower bounds sum to $0$, both must be attained, that is $A = 1/\beta^{\,n+1}$ and $B = -1/\beta^{\,n+1}$; in particular $\sum_{j>n}\bigl(\iota(c'_j)-\iota(c_j) + (\beta-1)\bigr)/\beta^{\,j+1} = 0$, a convergent series of nonnegative terms with sum $0$, so every term is $0$ and $\iota(c'_j) - \iota(c_j) = -(\beta-1)$ for every $j > n$.

An induction gives $x = \sum_{j<n} \iota(d_j)/\beta^{\,j+1} + r_n/\beta^{\,n}$ for every $n$: at $n = 0$ the sum is empty and $r_0/\beta^0 = x$; and from $r_n = (\iota(d_n) + r_{n+1})/\beta$, which is step 3.1 rearranged, one gets $r_n/\beta^{\,n} = \iota(d_n)/\beta^{\,n+1} + r_{n+1}/\beta^{\,n+1}$.

Since $0 \le r_n < 1$ and $\beta^{\,n} > 0$, we have $0 \le r_n/\beta^{\,n} \le 1/\beta^{\,n} = (1/\beta)^n$; as $0 < 1/\beta < 1$ the series $\sum (1/\beta)^k$ converges, so $(1/\beta)^n \to 0$, and the squeeze gives $r_n/\beta^{\,n} \to 0$.

That forces $c_j = b-1$ and $c'_j = 0$ for every $j > n$, since the difference of two digits attains $-(\beta-1)$ only at those values; so $(c_j)$ is terminal, contrary to hypothesis. Hence the two digit sequences agree, which is claim 2.

By step 5.1 the partial sums of $\sum_j \iota(d_j)/\beta^{\,j+1}$ equal $x - r_n/\beta^{\,n}$, which converges to $x$; so the series converges with sum $x$.

Applying step 5.1 and step 6.1 to the residue $r_J$ in place of $x$, whose recursion produces the digits $d_{J+i}$, gives $r_J = \sum_{i \ge 0} \iota(d_{J+i})/\beta^{\,i+1}$ for every $J$.

The constructed sequence is not terminal: if $d_j = b-1$ for every $j \ge J$, then by step 7.1 and the geometric series, $r_J = \sum_{i\ge0} (\beta-1)/\beta^{\,i+1} = \frac{\beta-1}{\beta}\cdot\frac{1}{1 - 1/\beta} = 1$, contradicting $r_J < 1$; here $\iota(b-1) = \beta - 1$ by [L1]. With step 6.1 this proves claim 1.

Claim 1 is step 6.1 with step 8.1 and claim 2 is step 5.3, so every $x \in [0,1)$ has exactly one non-terminal base-$b$ expansion.