Sum of Sequence as Summation of Difference of Adjacent Terms/Proof 2

Proof
From Abel's Lemma: Formulation 2, after renaming and reassigning variables:


 * $\ds \sum_{k \mathop = 1}^n a_k b_k = \sum_{k \mathop = 1}^{n - 1} \map {A_k} {a_k - a_{k + 1} } + A_n a_n$

where:
 * $\sequence a$ and $\sequence b$ are sequences in $\R$
 * $\ds A_n = \sum_{k \mathop = 1}^n {b_k}$ be the partial sum of $\sequence b$ from $1$ to $n$.

Let $\sequence b$ be defined as:
 * $\forall k: b_k = 1$

Thus:
 * $\ds A_n = \sum_{k \mathop = 1}^n 1 = n$

and so: