Chapter 8.3, Problem 89E

Remainder term Consider the geometric series $S={\displaystyle \underset{k=0}{\overset{\infty }{\sum }}{r}^k}$, which has the value 1/(1 − r) provided |r| < 1. Let $S_n={\displaystyle \underset{k=0}{\overset{n-1}{\sum }}{r}^k=\frac{1-r^n}{1-r}}$ be the sum of the first n terms.

The magnitude of the remainder R_n is the error in approximating S by S_n. Show that

$R_n=S-S_n=\frac{r^n}{1-r}.$