Abel's Lemma/Formulation 2/Proof 2

Proof
First note that:
 * $\ds A_{m - 1} = \sum_{i \mathop = m}^{m - 1} a_i = 0$

is a vacuous summation, as the upper index is smaller than the lower index.

Then we have:

Therefore:
 * $\ds \forall n \ge m: \sum_{k \mathop = m}^n a_k b_k = \sum_{k \mathop = m}^{n - 1} A_k \paren {b_k - b_{k + 1} } + A_n b_n$