At each step, there is a sum (green) and a remainder (red).[br][br]For the next step, the amount added to the sum and the amount left over are equal. [br][br]In algebra, if we have the sum[br] [math]S_k=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{2^k}[/math][br][br]Then[br][list=1][*][u]As k increases, the sum approaches 1:[/u] Given any finite number [i]x[/i], [i]no matter how small[/i], we can name the step when the red bar comes closer to B than [i]x[/i]. In other words, there exists no finite amount which will remain red, as long as we agree never to stop subdividing.[/*][*]Hence, we say that, in the limit as [math]k\longrightarrow\infty[/math], the complete sum S = 1.[/*][*][u]No tricks:[/u] By saying the sum (in the limit) S = 1, we respect the given information. The given length AB=1, and we divide in such a way that, [i]provided we agree never to stop dividing,[/i] no amount however small can be left over. [/*][*]if we were confronted with a finite number of terms:[br][math]S_k=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{2^k}[/math],[br]an infinite number of terms are discarded. Nevertheless, the sum of all these missing terms is [math]\frac{1}{2^k}[/math].[/*][/list][br]If we are given a sum whose value we [i]do not know[/i], we can use bisection as a frame of reference. For example, if each term in the sequence is less than half the term before it ... then if we agree to add [i]n[/i] terms, and then stop, the discarded amount (the error) is smaller than the last (n[i]th[/i]) term.