Abel's Lemma

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

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

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

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

Corollary
This lemma is usually reported as:
 * $$\sum_{i=0}^n a_i b_i = \sum_{i=0}^{n-1} A_i \left({b_i - b_{i+1}}\right) + A_n b_n$$

Its proof is trivial: set $$m = 0$$.

Proof
Proof by induction:

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

Basis for the Induction

 * First we consider $$P(m)$$.

Note that when $$n = m$$, we have that $$\sum_{i=m}^{n-1} A_i \left({b_i - b_{i+1}}\right) = 0$$ is a vacuous summation, as the upper index is smaller than the lower index.

We also have that $$A_m = \sum_{i=m}^m {a_i} = a_m$$.

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:
 * $$\sum_{i=m}^k a_i b_i = \sum_{i=m}^{k-1} A_i \left({b_i - b_{i+1}}\right) + A_k b_k$$

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

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

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