Abel's Lemma/Formulation 2/Corollary

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:

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

Proof

From Abel's Lemma: Formulation 2, we have:

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

The result follows by setting $m = 0$.

$\blacksquare$

Source of Name

This entry was named for Niels Henrik Abel.