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 $N\ge 1$ . For $N+1$ , we then have \begin{eqnarray*} \sum_{i=0}^{N+1} a_i b_i &=& \sum_{i=0}^{N} a_i b_i + a_{N+1} b_{N+1} \\ &=& \sum_{i=0}^{N-1}A_i(b_i-b_{i+1})+A_N b_N + a_{N+1} b_{N+1} \\ &=& \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}. \end{eqnarray*}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. .