Abel's Lemma/Formulation 2/Proof 1
Lemma
Let $\ds A_n = \sum_{i \mathop = m}^n {a_i}$ be the partial sum of $\sequence a$ from $m$ to $n$.
Then:
- $\ds \sum_{k \mathop = m}^n a_k b_k = \sum_{k \mathop = m}^{n - 1} A_k \paren {b_k - b_{k + 1} } + A_n b_n$
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:
\(\ds \sum_{k \mathop = m}^{r + 1} a_k b_k\) | \(=\) | \(\ds \sum_{k \mathop = m}^r a_k b_k + a_{r + 1} b_{r + 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{k \mathop = m}^{r - 1} A_k \paren {b_k - b_{k + 1} } + A_r b_r + a_{r + 1} b_{r + 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{k \mathop = m}^{r - 1} A_k b_k - \sum_{k \mathop = m}^{r - 1} A_k b_{k + 1} + A_r b_r + a_{r + 1} b_{r + 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{k \mathop = m}^r A_k b_k - \paren {\sum_{k \mathop = m}^r A_k b_{k + 1} - A_r b_{r + 1} } + a_{r + 1} b_{r + 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{k \mathop = m}^r A_k \paren {b_k - b_{k + 1} } + \paren {A_r + a_{r + 1} } b_{r + 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{k \mathop = m}^r A_k \paren {b_k - b_{k + 1} } + A_{r + 1} b_{r + 1}\) |
So $\map P n \implies \map P {r + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
- $\ds \forall n \ge m: \sum_{k \mathop = m}^n a_k b_k = \sum_{k \mathop = m}^{n - 1} A_k \paren {b_k - b_{k + 1} } + A_n b_n$
$\blacksquare$
Source of Name
This entry was named for Niels Henrik Abel.