Sum of infinite terms of a G.P.

Consider a G.P. of positive terms. The sum of the first $n$ terms is $S_n=\dfrac{a(r^n-1)}{r-1}=\dfrac{a(1-r^n)}{1-r}$, $r\ne1$. If $r>1$, $r^n$ grows without bound as $n\to\infty$, so the infinite sum cannot be found (does not exist). If $r<1$ (more precisely $|r|<1$), $r^n\to0$ as $n\to\infty$, so $S_n\to\dfrac{a}{1-r}$; this limiting value is called the sum to infinity, written $\sum_{r=1}^{\infty}t_r=\dfrac{a}{1-r}$.

Example: find $1+\dfrac12+\dfrac14+\dfrac18+\dfrac1{16}+\cdots$. Here $a=1, r=\dfrac12$ (so $r<1$), so the sum to infinity is $\dfrac{a}{1-r}=\dfrac{1}{1-1/2}=2$. The accompanying visual proof (Fig. 2.1) shows a $2\times1$ rectangle progressively tiled by rectangles of areas $1,\tfrac12,\tfrac14,\ldots$, which fill the big rectangle of area 2, confirming $1+\tfrac12+\tfrac14+\cdots=2$. …