proof of Abel’s lemma (by induction)

Proof. The proof is by induction. However, let usfirst recall that sum on the right side is apiece-wise defined function of the upper limit $N-1$ .In other words, if the upper limit is below the lowerlimit $0$ , 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 thatboth sides simplify to $a_{0}b_{0}+a_{1}b_{1}$ .Then, for the induction step, suppose that theclaim 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. $\\Box$ .