Sum of an Infinite Geometric Progression

Section 4 derived the sum of the first $n$ terms of a G.P. as $S_n = \dfrac{a(1-r^n)}{1-r}$ for $r\ne 1$. This section asks: does it make sense to sum infinitely many terms of a G.P., i.e. does $S_n$ approach a definite finite value as $n$ grows without bound?

The key behaviour of $r^n$. Everything depends on what happens to $r^n$ as $n\to\infty$:

• If $|r|<1$ (i.e. $-1<r<1$), then repeatedly multiplying a number of magnitude less than $1$ by itself makes its magnitude shrink further at every step, so $r^n \to 0$ as $n\to\infty$. For instance, with $r=\frac12$: $r^1=0.5,\ r^2=0.25,\ r^3=0.125,\ r^{10}\approx 0.00098$, clearly heading to $0$.
• If $|r|>1$, then $|r|^n$ grows without bound, so $S_n$ has no finite limit — the "sum" diverges.
• If $r=1$, $S_n=na\to\pm\infty$ (unless $a=0$); if $r=-1$, the terms are $a,-a,a,-a,\ldots$ and $S_n$ oscillates between $0$ and $a$ forever, never settling to one value.

The sum to infinity. Whenever $|r|<1$, since $r^n\to 0$, taking the limit of the finite-sum formula gives

This single formula replaces the finite-sum formula whenever a G.P. is summed "forever," and the condition $|r|<1$ is essential and must always be checked first — quoting this formula for a G.P. with $|r|\ge 1$ is meaningless, since no finite sum exists in that case.

Application: recurring decimals as an infinite G.P. A purely recurring decimal such as $0.\overline{d} = 0.dddd\ldots$ can be read as an infinite series

which is a G.P. with first term $a=\dfrac{d}{10}$ and common ratio $r=\dfrac{1}{10}$ (so $|r|<1$ always). By the boxed formula,

$0.\overline{d} = \frac{d/10}{1-1/10} = \frac{d/10}{9/10} = \frac{d}{9}.$ …