We illustrate the proof in the case that s-s_j consists of two kinds of terms: those that decrease by a factor c when j increases by 1, and those that decrease by a smaller factor, d (1 > c > d >0.). Similar argument applies in general.

We then have

s_j -s = Ac^j  + Bd ^j

and

t_j -s = s_j / (1 - c) - s_{j -1} c / (1 - c)-s = (s_j -s) / (1 - c) - s_{j -1} c / (1 - c) = B(d - c)d ^j-1

which decreases by a factor d as j increases by 1.

Thus our trick kills off the term in s_j that declines most slowly as j increases.