Abel's Lemma/Formulation 1/Corollary

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Corollary to Abel's Lemma: Formulation 1

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


Then:

$\ds \sum_{k \mathop = 1}^n a_k \paren {b_{k + 1} - b_k} = a_{n + 1} b_{n + 1} - a_1 b_1 - \sum_{k \mathop = 1}^n \paren {a_{k + 1} - a_k} b_{k + 1}$


Note that although proved for the general ring, this result is usually applied to one of the conventional number fields $\Z, \Q, \R$ and $\C$.


Proof

From Abel's Lemma: Formulation 1, we have:

$\ds \sum_{k \mathop = m}^n a_k \paren {b_{k + 1} - b_k} = a_{n + 1} b_{n + 1} - a_m b_m - \sum_{k \mathop = m}^n \paren {a_{k + 1} - a_k} b_{k + 1}$

The result follows by setting $m = 0$.

$\blacksquare$


Also reported as

Some sources give this as:

$\ds \sum_{k \mathop = 1}^n \paren {a_{k + 1} - a_k} b_k = a_{n + 1} b_{n + 1} - a_1 b_1 - \sum_{k \mathop = 1}^n a_{k + 1} \paren {b_{k + 1} - b_k}$

which is obtained from the main result by interchanging $a$ and $b$.


Others take the upper index to $n - 1$:

$\ds \sum_{k \mathop = 1}^{n - 1} \paren {a_{k + 1} - a_k} b_k = a_n b_n - a_1 b_1 - \sum_{k \mathop = 1}^{n - 1} a_{k + 1} \paren {b_{k + 1} - b_k}$


Source of Name

This entry was named for Niels Henrik Abel.


Sources