Abel's Lemma

Lemma

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

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

Then:

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

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:

$\displaystyle \sum_{i=0}^n a_i b_i = \sum_{i=0}^{n-1} A_i \left({b_i - b_{i+1}}\right) + A_n b_n$

Its proof is trivial: set $m = 0$ in the main proof.

Proof 1

Proof by induction:

For all $n \in \N$ where $n \ge m$, let $P \left({n}\right)$ be the proposition:

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

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:

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

Then we need to show:

$\displaystyle \sum_{i=m}^{k+1} a_i b_i = \sum_{i=m}^{k} A_i \left({b_i - b_{i+1}}\right) + A_{k+1} b_{k+1}$

where:

$\displaystyle A_{k+1} = \sum_{i=m}^{k+1} {a_i}$

Induction Step

This is our induction step:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \sum_{i=m}^{k+1} a_i b_i\)	\(=\)	\(\displaystyle \sum_{i=m}^k a_i b_i + a_{k+1} b_{k+1}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \sum_{i=m}^{k-1} A_i \left({b_i - b_{i+1} }\right) + A_k b_k + a_{k+1} b_{k+1}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \sum_{i=m}^{k-1} A_i b_i - \sum_{i=m}^{k-1} A_i b_{i+1} + A_k b_k + a_{k+1} b_{k+1}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \sum_{i=m}^{k} A_i b_i - \left({\sum_{i=m}^{k} A_i b_{i+1} - A_k b_{k+1} }\right) + a_{k+1} b_{k+1}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \sum_{i=m}^{k} A_i \left({b_i - b_{i+1} }\right) + \left({A_k + a_{k+1} }\right) b_{k+1}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \sum_{i=m}^{k} A_i \left({b_i - b_{i+1} }\right) + A_{k+1} b_{k+1}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore:

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

$\blacksquare$

Proof 2

First, note that : $a_k = A_k - A_{k-1}$.

Hence:

$\displaystyle\sum_{k=0}^n a_kb_k = \sum_{k=0}^n (A_k - A_{k-1}) b_k = \sum_{k=0}^n A_kb_k - \sum_{k=0}^n A_{k-1}kb_{k}$

Then, substitute $k$ for $i=k-1$ in the second sum so that:

$\displaystyle\sum_{k=0}^n A_{k-1}b_{k} =\sum_{i=0}^{n-1} A_{i} b_{i+1} = \sum_{k=0}^{n-1} A_{k} b_{k+1}$

We now have:

$\displaystyle\sum_{k=0}^n a_kb_k = \sum_{k=0}^n A_kb_k - \sum_{k=0}^{n-1} A_{k} b_{k+1} = \sum_{k=0}^{n-1} A_k(b_k-b_{k+1}) + A_nb_n$

$\blacksquare$

Source of Name

This entry was named for Niels Henrik Abel.