1.39 Proof

Proof. We show that the sum to $n$ terms is

$$s_{n}=a{{1-r^{n}}\over{1-r}}.$$

We take the $n^{th}$ partial sum, and then multiply by $r$, so

$$s_{n}=a+ar+ar^{2}+\dots+ar^{n-1}$$

$$rs_{n}={\quad}ar+ar^{2}+\dots+ar^{n-1}+ar^{n};$$

so by subtracting we obtain

$$(1-r)s_{n}=a-ar^{n};$$

hence when $r\neq 1$, we obtain

$$s_{n}=a{{1-r^{n}}\over{1-r}}.$$

Now for $-1<r<1$, we have $r^{n}\rightarrow 0$ as $n\rightarrow\infty$, so

$$s_{n}\rightarrow{{a}\over{1-r}}\qquad(n\rightarrow\infty)$$

and $s=a/(1-r).$