Abel's Lemma/Formulation 2/Corollary

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Lemma

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

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


Then:

$\displaystyle \sum_{k \mathop = 0}^n a_k b_k = \sum_{k \mathop = 0}^{n - 1} A_k \left({b_k - b_{k + 1} }\right) + A_n b_n$


Proof

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

$\displaystyle \sum_{k \mathop = m}^n a_k b_k = \sum_{k \mathop = m}^{n - 1} A_k \left({b_k - b_{k + 1}}\right) + A_n b_n$

The result follows by setting $m = 0$.

$\blacksquare$


Source of Name

This entry was named for Niels Henrik Abel.