16.3 Geometric Series¶

Consider the series:

$\sum_{k=0}^n c^k$

for some $c \neq 0$ and $c \neq 1$ .

We can calculate the sum by simply multiplying the series by $(1-c)$ :

$(1 + c + c^2 + c^3 + \dots c^n)(1-c) = 1 - c^{n+1}$

Dividing both sides by $(1-c)$ yields:

$\sum_{k=0}^n c^k = \frac{1}{1-c} - \frac{c^{n+1}}{1-c}$

What happens as we increase $n \rightarrow \infty$ ?

If $c > 1$ , then the term on the right will blow up and also increase off to infinity. This isn’t very interesting, so let’s ignore it for now.
If $0 < c < 1$ , then the term on the right will get smaller and smaller and eventually become 0.

This second case is most interesting, because it gives us:

$\sum_{k=0}^\infty c^k = \frac{1}{1-c} \text{ when } 0 < c < 1$

Note that the $\infty$ symbol is not a number. In this context, it is the idea that $k$ keeps increasing on and on forever.

As an example, suppose $c = \frac{1}{2}$ :

$\sum_{k=0}^\infty \left(\frac{1}{2}\right)^k = \frac{1}{1-\frac{1}{2}} = \frac{1}{\frac{1}{2}} = 2$