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.

If $\abs{x} < 1$, then the geometric series of $x$ converges and

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

If $\abs{x} \geq 1$ then the geometric series of $x$ diverges.

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

\[ \begin{aligned} x \cdot s_n &= x\cdot (x^1 + x^2 + \cdots + x^n) \\ &= x^2 + x^3 + \cdots + x^{n+1} \\ &= s_n - x + x^{n+1}. \end{aligned} \]

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 $\abs{x} < 1$, 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 $\abs{x} < 1$.

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 $\abs{x} > 1$. ↩︎