Geometric series

Infinite series: Common series

Geometric series

In the chapter Limits of sequences we studied geometric sequences:

A geometric sequence is a sequence for which there exists a constant $z$ such that $a_{n+1} = z\cdot a_n$ for all $n$ .

This implies $a_n = a_0 r^n$ . The series corresponding to a geometric sequence is aptly called a geometric series. In the last exercise we have already seen a geometric series:

In this figure, $s_n = \frac 12 + \left(\frac{1}{2}\right)^2 + \dots + \left(\frac{1}{2}\right)^n$ is displayed for $n$ between $1$ and $10$ .

What is the value of the series

$\sum_{n=1}^\infty \left(\frac{1}{2}\right)^n \text?$

The part of the circle that's white becomes twice as small in each step. This means that the area of the white portion of the circle converges to zero, so the area of the blue part converges towards $1$ . The answer therefore is

$\sum_{n=1}^\infty \left(\frac{1}{2}\right)^n = 1\text.$

New example

If we add $a_0 = \left(\frac{1}{2}\right)^0 = 1$ at the beginning of the sequence we obtain $\sum_{n=0}^\infty \left( \frac{1}{2} \right)^n = 2$ .
The following formula generalises this result:

For $|r|<1$ we have

$\sum_{n=0}^\infty r^n = \frac{1}{1-r}\text.$

By expanding the brackets we see that

$\left(1-r\right)\left( 1 + r + r^2 + \dots + r^n\right) = 1 - r^{n+1}\text,$ since all other terms cancel. This means that the partial sums equal

$\sum_{k=0}^n r^k = \frac{1-r^{n+1}}{1-r}\text,$ and because $|r|<1$ we know that $\lim_{n\to\infty} r^{n+1}=0$ , so the numerator converges to $1$ . The partial sums therefore converge to $\frac{1}{1-r}$ .