Geometric Series

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.