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. $\\Box$