Abel's Lemma/Formulation 2/Proof 1

Proof
Proof by induction:

For all $n \in \N$ where $n \ge m$, let $\map P n$ be the proposition:
 * $\ds \sum_{k \mathop = m}^n a_k b_k = \sum_{k \mathop = m}^{n - 1} A_k \paren {b_k - b_{k + 1} } + A_n b_n$

Basis for the Induction
First consider $\map P m$.

When $n = m$, we have that:
 * $\ds \sum_{k \mathop = m}^{n - 1} A_k \paren {b_k - b_{k + 1} } = 0$

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

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

Thus we see that $\map P m$ 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 $\map P r$ is true, where $r \ge m$, then it logically follows that $\map P {r + 1}$ is true.

So this is our induction hypothesis:
 * $\ds \sum_{i \mathop = m}^r a_k b_k = \sum_{k \mathop = m}^{r - 1} A_k \paren {b_k - b_{k + 1} } + A_r b_r$

Then we need to show:
 * $\ds \sum_{k \mathop = m}^{r + 1} a_k b_k = \sum_{k \mathop = m}^r A_k \paren {b_k - b_{k + 1} } + A_{r + 1} b_{r + 1}$

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

Induction Step
This is our induction step:

So $\map P n \implies \map P {r + 1}$ and the result follows by the Principle of Mathematical Induction.

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