Abel lemma

$\displaystyle \sum_{i=0}^n a_ib_i=\sum_{i=0}^{n-1} A_i(b_i-b_{i+1})+A_nb_n,$

(1)

where, $A_i=\sum_{k=0}^i a_k$ . Sequences $\{a_i\}$ , $\{b_i\}$ , $i=0,\dots, n$ , are real or complex one.

Proof

We consider the expansion of the sum

$\displaystyle \sum_{i=0}^n A_i(b_i-b_{i+1})$

on two different forms, namely:

On the short way.

$\displaystyle \sum_{i=0}^n A_i(b_i-b_{i+1})=\sum_{i=0}^{n-1} A_i(b_i-b_{i+1})+A_nb_n-A_nb_{n+1}.$ (2)
On the long way.

$\displaystyle \sum_{i=0}^n A_i(b_i-b_{i+1})=\sum_{i=0}^n A_ib_i-\sum_{i=0}^n A_ib_{i+1}= \sum_{i=0}^n A_ib_i-\sum_{i=1}^{n+1} A_{i-1}b_i=$

$\displaystyle A_0 b_0+\sum_{i=1}^n (A_{i-1}+a_i)b_i-\sum_{i=1}^n A_{i-1}b_i-A_nb_{n+1}=\sum_{i=0}^n a_ib_i-A_nb_{n+1},$

(3)

where a simplification has been performed. Notice that

. By equating (2), (3), the last two terms cancel, ¹ and then, (1) follows. $\Box$

Footnotes

...¹: Without loss of generality, $b_{n+1}$ may be assumed finite. Indeed we don't need $b_{n+1}$ , but the proof is a couple lines larger. It is left as an exercise.

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Cross-references: lines, finite, without loss of generality, terms, sum, complex, real, sequences

This is version 4 of proof of Abel lemma (by expansion), born on 2007-08-12, modified 2007-08-14.
Object id is 9856, canonical name is ProofOfAbelLemmaByExpansion.
Accessed 481 times total.

Classification:

AMS MSC:

40A05 (Sequences, series, summability :: Convergence and divergence of infinite limiting processes :: Convergence and divergence of series and sequences)

Pending Errata and Addenda

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