Abel's Lemma/Formulation 2/Proof 1
Lemma
Let $\displaystyle A_n = \sum_{i \mathop = m}^n {a_i}$ be the partial sum of $\sequence a$ from $m$ to $n$.
Then:
- $\displaystyle \sum_{k \mathop = m}^n a_k b_k = \sum_{k \mathop = m}^{n - 1} A_k \paren {b_k - b_{k + 1} } + A_n b_n$
Proof
Proof by induction:
For all $n \in \N$ where $n \ge m$, let $P \left({n}\right)$ be the proposition:
- $\displaystyle \sum_{k \mathop = m}^n a_k b_k = \sum_{k \mathop = m}^{n - 1} A_k \left({b_k - b_{k + 1} }\right) + A_n b_n$
Basis for the Induction
First consider $P(m)$.
When $n = m$, we have that:
- $\displaystyle \sum_{k \mathop = m}^{n - 1} A_k \left({b_k - b_{k + 1} }\right) = 0$
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 $P \left({r}\right)$ is true, where $r \ge m$, then it logically follows that $P \left({r + 1}\right)$ is true.
So this is our induction hypothesis:
- $\displaystyle \sum_{i \mathop = m}^r a_k b_k = \sum_{k \mathop = m}^{r - 1} A_k \left({b_k - b_{k + 1} }\right) + A_r b_r$
Then we need to show:
- $\displaystyle \sum_{k \mathop = m}^{r + 1} a_k b_k = \sum_{k \mathop = m}^r A_k \left({b_k - b_{k + 1} }\right) + A_{r + 1} b_{r + 1}$
where:
- $\displaystyle A_{r + 1} = \sum_{i \mathop = m}^{r + 1} {a_i}$
Induction Step
This is our induction step:
\(\ds \sum_{k \mathop = m}^{r + 1} a_k b_k\) | \(=\) | \(\ds \sum_{k \mathop = m}^r a_k b_k + a_{r + 1} b_{r + 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{k \mathop = m}^{r - 1} A_k \left({b_k - b_{k + 1} }\right) + A_r b_r + a_{r + 1} b_{r + 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{k \mathop = m}^{r - 1} A_k b_k - \sum_{k \mathop = m}^{r - 1} A_k b_{k + 1} + A_r b_r + a_{r + 1} b_{r + 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{k \mathop = m}^r A_k b_k - \left({\sum_{k \mathop = m}^r A_k b_{k + 1} - A_r b_{r + 1} }\right) + a_{r + 1} b_{r + 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{k \mathop = m}^r A_k \left({b_k - b_{k + 1} }\right) + \left({A_r + a_{r + 1} }\right) b_{r + 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{k \mathop = m}^r A_k \left({b_k - b_{k + 1} }\right) + A_{r + 1} b_{r + 1}\) |
So $P \left({r}\right) \implies P \left({r + 1}\right)$ and the result follows by the Principle of Mathematical Induction.
Therefore:
- $\displaystyle \forall n \ge m: \sum_{k \mathop = m}^n a_k b_k = \sum_{k \mathop = m}^{n - 1} A_k \left({b_k - b_{k + 1} }\right) + A_n b_n$
$\blacksquare$
Source of Name
This entry was named for Niels Henrik Abel.