Abel's Lemma/Formulation 2/Proof 1

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
Proof by induction:

For all $n \in \N$ where $n \ge m$, let $P \left({n}\right)$ be the proposition:
 * $\displaystyle \sum_{i \mathop = m}^n a_i b_i = \sum_{i \mathop = m}^{n-1} A_i \left({b_i - b_{i+1}}\right) + A_n b_n$

Basis for the Induction
First consider $P(m)$.

When $n = m$, we have that:
 * $\displaystyle \sum_{i \mathop = m}^{n-1} A_i \left({b_i - b_{i+1}}\right) = 0$

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

We also have that:
 * $\displaystyle A_m = \sum_{i \mathop = m}^m {a_i} = a_m$

Thus we see that $P(m)$ is true, as this just says:
 * $a_m b_m = 0 + A_m b_m = a_m b_m$

which is clearly true.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge m$, then it logically follows that $P \left({k+1}\right)$ is true.

So this is our induction hypothesis:
 * $\displaystyle \sum_{i \mathop = m}^k a_i b_i = \sum_{i \mathop = m}^{k-1} A_i \left({b_i - b_{i+1}}\right) + A_k b_k$

Then we need to show:
 * $\displaystyle \sum_{i \mathop = m}^{k+1} a_i b_i = \sum_{i \mathop = m}^{k} A_i \left({b_i - b_{i+1}}\right) + A_{k+1} b_{k+1}$

where:
 * $\displaystyle A_{k+1} = \sum_{i \mathop = m}^{k+1} {a_i}$

Induction Step
This is our induction step:

So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction.

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