Geometric Series

Have you ever taken out a loan, or thought about how much interest you would pay if you did? Then you were actually thinking about Geometric series, which can be used to compute the APR of a loan.

Definition of a Geometric Series

How do you know if a series is geometric or not? It has to do with the sequence you make it from. If the sequence that makes up the series is geometric, then the series is geometric. Remember that a geometric sequence is where you get each new term in the sequence by multiplying the previous one by a constant. So the sequence will have the form $\left\{a,ar,a{r}^2,a{r}^3,\dots \right\}$ where $r\ne 0$.

A geometric series is a series that is formed by summing the terms from a geometric sequence.

Formula for a Geometric Series

It is handy to look at the summation notation of a geometric series. The geometric series made from a geometric sequence looks like

$\sum _{n=1}^{\infty }a{r}^{n-1}$

where $a$ and id="2938751" role="math" $r\ne 0$ are constant real numbers. Just like with a geometric sequence, $r$ is called the common ratio.

Partial Sums of a Geometric Series

If you are taking out a loan, you certainly don't want to make infinitely many payments! So it can help to have a formula for the partial sums of a geometric series. The n^th partial sum is

$s_n=a+\hspace{0.17em}ar+a{r}^2+\dots +a{r}^{n-1}$.

Notice that this is essentially an (n-1)^st degree polynomial where $r$ is the variable.

What happens if you multiply both sides by $r$? Then you get

$r{s}_n=ar+\hspace{0.17em}a{r}^2+a{r}^3+\dots +a{r}^n$.

So if you subtract the two equations, you get

$\begin{array}{ccc}{s}_n-r{s}_n& =& \left(a+r+r^2+\dots +\hspace{0.17em}a{r}^{n-1}\right)-\left(ar+a{r}^2+\dots +a{r}^n\right)\\ s_n-r{s}_n& =& a-a{r}^n\end{array}$

which is awfully nice, because then you can easily solve for $s_n$ as long as $r\ne 1$ to get

$s_n=\frac{a-a{r}^n}{1-r}=\frac{a\left(1-r^n\right)}{1-r}$.

Convergence of a Geometric Series

Once you have a nice formula for the partial sums of a series, you can look at the limit to see when it converges. So let's look at some values of $r$ to see when

$\underset{n\to \infty }{\lim }{s}_n=\underset{n\to \infty }{\lim }\frac{a\left(1-r^n\right)}{1-r}$

exists. Doing some algebra,

$$\begin{gathered} \underset{n\to \infty }{\lim }{s}_n=\underset{n\to \infty }{\lim }\frac{a\left(1-r^n\right)}{1-r} \\ =\underset{n\to \infty }{\lim }\left[\frac{a}{1-r}-\frac{a{r}^n}{1-r}\right]. \end{gathered}$$

The first part of the limit doesn't depend on $n$, but you certainly need to be sure that $r\ne 1$ so you aren't dividing by zero. Factoring out the constants from the second part, you have

$$\begin{gathered} \underset{n\to \infty }{\lim }\left[\frac{a}{1-r}-\frac{a{r}^n}{1-r}\right]=\frac{a}{1-r}-\frac{a}{1-r}\underset{n\to \infty }{\lim }{r}^n \\ =\left(\frac{a}{1-r}\right)\left[1-\underset{n\to \infty }{\lim }{r}^n\right]. \end{gathered}$$

So you can see that if

$\underset{n\to \infty }{\lim }{r}^n$

exists, then the limit of the series will exist as well.

That limit exists when $r$ is between $-1$ and $1$. But you still have to be a little careful, because it isn't a geometric sequence if $r=0$, and you can't actually use $r=1$ (because that gave you division by zero) or $r=-1$ (because the sequence with $r=-1$ doesn't converge).

Sum of a Geometric Series

Geometric series are especially nice because you can say when they converge, and exactly what they converge to. From the previous discussion, a geometric series converges when $-1<r<1$ and diverges otherwise. When the geometric series converges, taking the limit of the partial sums gives you:

$\sum _{n=1}^{\infty }a{r}^{n-1}=\frac{a}{1-r}$.

Examples with Geometric Series

Let's look at some examples and see what geometric series can tell you.