proof of Abel’s lemma (by induction)

Proof. The proof is by induction. However, let us first recall that sum on the right side is a piece-wise defined function of the upper limit $N-1$ . In other words, if the upper limit is below the lower limit $0$ , the sum is identically set to zero. Otherwise, it is an ordinary sum. We therefore need to manually check the first two cases. For the trivial case $N=0$ , both sides equal to $a_{0}b_{0}$ . Also, for $N=1$ (when the sum is a normal sum), it is easy to verify that both sides simplify to $a_{0}b_{0}+a_{1}b_{1}$ . Then, for the induction step, suppose that the claim holds for some $N\geq 1$ . For $N+1$ , we then have

$\displaystyle\sum_{i=0}^{N+1}a_{i}b_{i}$	$\displaystyle=$	$\displaystyle\sum_{i=0}^{N}a_{i}b_{i}+a_{N+1}b_{N+1}$
	$\displaystyle=$	$\displaystyle\sum_{i=0}^{N-1}A_{i}(b_{i}-b_{i+1})+A_{N}b_{N}+a_{N+1}b_{N+1}$
	$\displaystyle=$	$\displaystyle\sum_{i=0}^{N}A_{i}(b_{i}-b_{i+1})-A_{N}(b_{N}-b_{N+1})+A_{N}b_{N% }+a_{N+1}b_{N+1}.$

Since $-A_{N}(b_{N}-b_{N+1})+A_{N}b_{N}+a_{N+1}b_{N+1}=A_{N+1}b_{N+1}$ , the claim follows. $\Box$ .

Title	proof of Abel’s lemma (by induction)
Canonical name	ProofOfAbelsLemmabyInduction
Date of creation	2013-03-22 13:38:04
Last modified on	2013-03-22 13:38:04
Owner	mathcam (2727)
Last modified by	mathcam (2727)
Numerical id	9
Author	mathcam (2727)
Entry type	Proof
Classification	msc 40A05