For every ordinal $α$, let $r_α$ be a positive real number. I then define $G$ recursively as $$ G(α) = \sup\{G(β)+r_β: β\in α\}. $$ My intention is to prove that $$ G(α) = \sum_{i\in α}r_i .$$ Here is my reasoning. By inductive hypothesis, $$ G(α) = \sup\left\{ \sum_{i\in β}r_{i} + r_β: β\in α\right\}$$ and by definition, $$ \sum_{i\in α}r_i = \sup\left\{ \sum_{i\in n}r_{f(i)}: n\in ω \land f\colon n+1 \to α \text{ increasing}\right\}$$ But $\sum_{i\in n}r_{f(i)} \le \sum_{i\in f(n)}r_{i}+r_{f(n)}$ and thus $G(α) \ge \sum_{i\in α}r_i$. The reverse inequality holds too, as $\sum_{i\in β+1}r_i \le \sum_{i\in α}r_i$ for every $β\in α$.
The purpose of this was to justify identities such as $\sum_{i∈α} (r_{i+1}-r_i) = \sup_{i∈α}r_{i+1} - r_0$, where $r_0,r_1,\dots $ is an increasing $α$-sequence of real numbers such that $\forall β \in α\; \sup_{i∈β}r_{i+1} = r_β$.
Are there any flaws in what I've written?