# 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) ProofOfAbelsLemmabyInduction 2013-03-22 13:38:04 2013-03-22 13:38:04 mathcam (2727) mathcam (2727) 9 mathcam (2727) Proof msc 40A05