Abel Summation

For $k=1,2,\dots n$, denote $\sum_{i=1}^k=S_k$, and $S_0=0$, then we have

$$\sum_{i=1}^b a_ib_i = S_nb_n + \sum_{i=1}^{n-1} S_i(b_i-b_{i+1})$$

Proof. Since $a_i=S_i-S_{i-1}$, this can be proved by simply expanding the expression. $\blacksquare$