AlgebraAMC 10

Geometric Series

S_n = a\cdot\frac{1-r^n}{1-r}, \qquad S_\infty = \frac{a}{1-r}\ \ (|r|<1)

First term $a$ , common ratio $r$ , $n$ terms.

What it says

A geometric series multiplies by the same ratio $r$ each step. The finite formula adds the first $n$ terms; when $|r| < 1$ the terms shrink toward zero and the whole infinite series adds up to $\frac{a}{1-r}$ . Repeating decimals and many probability problems are geometric series in disguise.

Show proof

Let $S = a + ar + \cdots + ar^{n-1}$ . Then $rS = ar + \cdots + ar^n$ , and subtracting gives $S - rS = a - ar^n$ , so $S = a\frac{1 - r^n}{1 - r}$ . If $|r| < 1$ , $r^n \to 0$ as $n \to \infty$ , leaving $\frac{a}{1-r}$ .

When to use it

Each term is a constant multiple of the one before it, or an infinite repeating process needs summing.

Example

Evaluate $1 + \frac13 + \frac19 + \frac1{27} + \cdots$ .

Show solution

$\frac{1}{1 - 1/3} = \frac32$ . Answer: $\frac{3}{2}$