proof of Abel lemma (by expansion)

1 Abel lemma

\sum_{i=0}^{n}a_{i}b_{i}=\sum_{i=0}^{n-1}A_{i}(b_{i}-b_{i+1})+A_{n}b_{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.

2 Proof

We consider the expansion of the sum

\sum_{i=0}^{n}A_{i}(b_{i}-b_{i+1})

on two different forms, namely:

1.

On the short way.

$\sum_{i=0}^{n}A_{i}(b_{i}-b_{i+1})=\sum_{i=0}^{n-1}A_{i}(b_{i}-b_{i+1})+A_{n}b% _{n}-A_{n}b_{n+1}.$ (2)
2.

On the long way.

\sum_{i=0}^{n}A_{i}(b_{i}-b_{i+1})=\sum_{i=0}^{n}A_{i}b_{i}-\sum_{i=0}^{n}A_{i% }b_{i+1}=\sum_{i=0}^{n}A_{i}b_{i}-\sum_{i=1}^{n+1}A_{i-1}b_{i}=

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_{n}% b_{n+1}=\sum_{i=0}^{n}a_{i}b_{i}-A_{n}b_{n+1},

(3)

where a simplification has been performed. Notice that $A_{0}=a_{0}$ . By equating (2), (3), the last two terms cancel, ¹¹Without 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. and then, (1) follows. $\Box$

Title	proof of Abel lemma (by expansion)
Canonical name	ProofOfAbelLemmabyExpansion
Date of creation	2013-03-22 17:28:14
Last modified on	2013-03-22 17:28:14
Owner	perucho (2192)
Last modified by	perucho (2192)
Numerical id	7
Author	perucho (2192)
Entry type	Proof
Classification	msc 40A05