Geometric Series

Why

It is believable that $1 / 2, 1 / 4, 1 / 8, \dots$ has a convergent series. And likewise with $1 / 3, 1 / 9, 1 / 27, \dots$ What of $a_{k} = x^{k}$ for $x \in R$ .

Definition

Let $x \in R$ . The geometric series of $x$ is the series of the sequence $(a_{k})$ defined by $a_{k} = x^{k}$ .

Characterization of convergence

Does the geometric series of $x$ converge? In other words, does $(s_{n})$ defined by $s_{n} = \sum_{k = 1}^{n} x^{k}$ have a limit.

For $x = 1$ and $x = - 1$ , we have seen (see Real Series) that the series diverges. However for the cases $x = 1 / 2$ and $x = 1 / 3$ the geometric series converges.

| x | < 1

, then the geometric series of

x

converges and

$lim_{n \to \infty} \sum_{k = 1}^{n} x^{k} = \frac{x}{1 - x}$

| x | \geq 1

then the geometric series of

x

diverges.

Define $s_{n} = \sum_{k = 1}^{n} x^{k}$ . Then

$\begin{aligned} x \cdot s_{n} & = x \cdot (x^{1} + x^{2} + \dots + x^{n}) \\ = x^{2} + x^{3} + \dots + x^{n + 1} \\ = s_{n} - x + x^{n + 1} . \end{aligned}$

From which we deduce,

s_{n} (1 - x) = x (1 - x^{n})

. If

x \neq 1

, then

$s_{n} = \frac{1}{1 - x} (1 - x^{n})$

If $| x | < 1$ , then using the algebra of limits (see Real Limit Algebra) we deduce

$lim_{n \to \infty} \frac{1}{1 - x} (1 - x^{n}) = \frac{1}{1 - x} (1 - 0) = \frac{1}{1 - x},$

since

lim_{k \to \infty} x^{k} = 0

for

| x | < 1

If $x = 1$ or $x = - 1$ , then we have seen that $(s_{n})$ diverges.¹

Future editions will include the trivial account about the case $| x | > 1$ . ↩︎