Abel's Lemma/Formulation 2/Proof 2

Lemma
Let $\left \langle {a} \right \rangle$ and $\left \langle {b} \right \rangle$ be sequences in an arbitrary ring $R$.

Let $\displaystyle A_n = \sum_{i \mathop = m}^n {a_i}$ be the partial sum of $\left \langle {a} \right \rangle$ from $m$ to $n$.

Then:
 * $\displaystyle \sum_{i \mathop = m}^n a_i b_i = \sum_{i \mathop = m}^{n-1} A_i \left({b_i - b_{i+1}}\right) + A_n b_n$

Note that although proved for the general ring, this result is usually applied to one of the conventional number fields $\Z, \Q, \R$ and $\C$.

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

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

Then we have:

Therefore:
 * $\displaystyle \forall n \ge m: \sum_{i \mathop = m}^n a_i b_i = \sum_{i \mathop = m}^{n-1} A_i \left({b_i - b_{i+1}}\right) + A_n b_n$