# Abel's Lemma/Formulation 2

## Lemma

Let $\sequence a$ and $\sequence b$ be sequences in an arbitrary ring $R$.

Let $\ds A_n = \sum_{i \mathop = m}^n {a_i}$ be the partial sum of $\sequence a$ from $m$ to $n$.

Then:

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

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

$\ds \sum_{k \mathop = 0}^n a_k b_k = \sum_{k \mathop = 0}^{n - 1} A_k \paren {b_k - b_{k + 1} } + A_n b_n$

## Proof 1

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:

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

$\blacksquare$

## Proof 2

First note that:

$\ds A_{m - 1} = \sum_{i \mathop = m}^{m - 1} a_i = 0$

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

Then we have:

 $\ds \sum_{k \mathop = m}^n a_k b_k$ $=$ $\ds \sum_{k \mathop = m}^n \paren {A_k - A_{k - 1} } b_k$ $\ds$ $=$ $\ds \sum_{k \mathop = m}^n A_k b_k - \sum_{k \mathop = m}^n A_{k - 1} b_k$ $\ds$ $=$ $\ds \sum_{k \mathop = m}^n A_k b_k - \sum_{k \mathop = m-1}^{n - 1} A_k b_{k + 1}$ $\ds$ $=$ $\ds \sum_{k \mathop = m}^{n - 1} A_k \paren {b_k - b_{k + 1} } + A_n b_n - A_{m - 1} b_m$ $\ds$ $=$ $\ds \sum_{k \mathop = m}^{n - 1} A_k \paren {b_k - b_{k + 1} } + A_n b_n - 0$

Therefore:

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

$\blacksquare$

## Also known as

Abel's Lemma is also known as:

Abel's transformation
Abel's partial summation formula
the technique of Summation by Parts.

## Source of Name

This entry was named for Niels Henrik Abel.