What happens if a geometric series never stops? Sometimes the total grows forever. But when the terms shrink fast enough, the partial sums settle on a finite number.
Key idea
If $|r| < 1$, $$a + ar + ar^2 + \cdots = \frac{a}{1 - r}.$$ If $|r| \ge 1$ (and $a \ne 0$), the series has no sum: the partial sums grow without bound or keep swinging.
Why it works
The partial sums are $S_n = \frac{a(1 - r^n)}{1 - r}$. When $|r| < 1$, $r^n$ shrinks toward 0 as $n$ grows ($\left(\frac{1}{2}\right)^{20}$ is about one in a million), so $S_n$ approaches $\frac{a}{1 - r}$. When $|r| \ge 1$, $r^n$ does not shrink, so the partial sums never settle. Walking halfway to a wall, then half the remaining distance, and so on, covers $\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots = 1$ wall-length in the limit.
Worked examples
Worked example: A convergent series
Find $12 + 4 + \frac{4}{3} + \cdots$
- $r = \frac{4}{12} = \frac{1}{3}$, and $|r| < 1$.
- $S = \frac{12}{1 - \frac{1}{3}} = \frac{12}{\frac{2}{3}} = 18$.
Answer: $18$
Worked example: An alternating series
Find $8 - 4 + 2 - 1 + \cdots$
- $r = -\frac{1}{2}$.
- $S = \frac{8}{1 + \frac{1}{2}} = \frac{16}{3}$.
Answer: $\frac{16}{3}$
Worked example: A repeating decimal
Write $0.\overline{27}$ as a fraction.
- $0.272727\ldots = \frac{27}{100} + \frac{27}{10000} + \cdots$, with $a = \frac{27}{100}$ and $r = \frac{1}{100}$.
- $S = \frac{27/100}{99/100} = \frac{27}{99} = \frac{3}{11}$.
Answer: $\frac{3}{11}$
The sum of an infinite series means the number its partial sums approach. It is a limit, an idea you will meet again in calculus, where series like these are used to compute values of $e^x$ and $\sin x$.
Common mistake
Using $\frac{a}{1 - r}$ when $|r| \ge 1$. For $2 + 4 + 8 + \cdots$ the formula gives $\frac{2}{1 - 2} = -2$, a negative "sum" of positive numbers. That is nonsense: the partial sums $2, 6, 14, 30, \ldots$ grow without bound. Check $|r|$ first.