Introduction to Series: Learn It 3

Geometric Series

A geometric series is any series that we can write in the form:

[latex]a+ar+a{r}^{2}+a{r}^{3}+\cdots =\displaystyle\sum _{n=1}^{\infty }a{r}^{n - 1}[/latex].

Because the ratio of each term in this series to the previous term is [latex]r[/latex], the number [latex]r[/latex] is called the common ratio. We refer to [latex]a[/latex] as the initial term because it is the first term in the series.

components of a geometric series

The number [latex]r[/latex] is called the common ratio because it’s the ratio between consecutive terms. The value [latex]a[/latex] is the initial term since it’s the first term in the series.

Let’s look at a concrete example. The series:

[latex]\displaystyle\sum _{n=1}^{\infty }{\left(\frac{1}{2}\right)}^{n - 1}=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots[/latex]

is a geometric series with initial term [latex]a=1[/latex] and common ratio [latex]r=\frac{1}{2}[/latex].

When Does a Geometric Series Converge?

The key question is: when does a geometric series actually converge to a finite value? To answer this, we need to examine the partial sums. Consider the geometric series [latex]\displaystyle\sum _{n=1}^{\infty }a{r}^{n - 1}[/latex] when [latex]a>0[/latex].

The sequence of partial sums [latex]\left\{{S}_{k}\right\}[/latex] is:

[latex]{S}_{k}=\displaystyle\sum _{n=1}^{k}a{r}^{n - 1}=a+ar+a{r}^{2}+\cdots +a{r}^{k - 1}[/latex].

 

Case 1: When [latex]r=1[/latex]

When [latex]r=1[/latex], our partial sum becomes:

[latex]{S}_{k}=a+a\left(1\right)+a{\left(1\right)}^{2}+\cdots +a{\left(1\right)}^{k - 1}=ak[/latex].

 

Since [latex]a>0[/latex], we have [latex]ak\to \infty[/latex] as [latex]k\to \infty[/latex]. The sequence of partial sums is unbounded, so the series diverges when [latex]r=1[/latex].

Case 2: When [latex]r\ne 1[/latex]

For [latex]r\ne 1[/latex], to find the limit of [latex]\left\{{S}_{k}\right\}[/latex], multiply the geometric series general equation by [latex]1-r[/latex].

[latex]\begin{array}{cc}\hfill \left(1-r\right){S}_{k}& =a\left(1-r\right)\left(1+r+{r}^{2}+{r}^{3}+\cdots +{r}^{k - 1}\right)\hfill \\ & =a\left[\left(1+r+{r}^{2}+{r}^{3}+\cdots +{r}^{k - 1}\right)-\left(r+{r}^{2}+{r}^{3}+\cdots +{r}^{k}\right)\right]\hfill \\ & =a\left(1-{r}^{k}\right).\hfill \end{array}[/latex]

 

Notice how most terms cancel out—this is the key insight! Therefore:

[latex]{S}_{k}=\frac{a\left(1-{r}^{k}\right)}{1-r}\text{for }r\ne 1[/latex].

From our discussion in the previous section, we know that the geometric sequence [latex]{r}^{k}\to 0[/latex] if [latex]|r|<1[/latex] and that [latex]{r}^{k}[/latex] diverges if [latex]|r|>1[/latex] or [latex]r= \pm{1}[/latex].

Using this knowledge:

  • For [latex]|r|<1[/latex]: [latex]{S}_{k}\to \frac{a}{1-r}[/latex], so [latex]\displaystyle\sum _{n=1}^{\infty }a{r}^{n - 1}=\frac{a}{1-r}[/latex]
  • For [latex]|r|\ge 1[/latex]: [latex]{S}_{k}[/latex] diverges, so [latex]\displaystyle\sum _{n=1}^{\infty }a{r}^{n - 1}[/latex] diverges.

 

convergence of geometric series

A geometric series is a series of the form

[latex]\displaystyle\sum _{n=1}^{\infty }a{r}^{n - 1}=a+ar+a{r}^{2}+a{r}^{3}+\cdots[/latex].

 

If [latex]|r|<1[/latex], the series converges, and

[latex]\displaystyle\sum _{n=1}^{\infty }a{r}^{n - 1}=\frac{a}{1-r}\text{ for }|r|<1[/latex].

 

If [latex]|r|\ge 1[/latex], the series diverges.

Recognizing Geometric Series in Different Forms

Geometric series don’t always appear in the standard form. You might encounter series where the index starts at a different value or the exponent has a different linear expression. The key is recognizing when you can rewrite the series in geometric form.

Identifying Geometric Series: Look for a constant ratio between consecutive terms. If you find one, you likely have a geometric series that can be rewritten in standard form.

Let’s work through an example.

Consider the series:

[latex]\displaystyle\sum _{n=0}^{\infty }{\left(\frac{2}{3}\right)}^{n+2}[/latex]

To see if this is geometric, write out the first several terms:

[latex]\begin{array}{cc}\hfill \displaystyle\sum _{n=0}^{\infty }{\left(\frac{2}{3}\right)}^{n+2}& ={\left(\frac{2}{3}\right)}^{2}+{\left(\frac{2}{3}\right)}^{3}+{\left(\frac{2}{3}\right)}^{4}+\cdots \hfill \\ & =\frac{4}{9}+\frac{4}{9}\cdot \left(\frac{2}{3}\right)+\frac{4}{9}\cdot {\left(\frac{2}{3}\right)}^{2}+\cdots .\hfill \end{array}[/latex]

We see that the initial term is [latex]a=\frac{4}{9}[/latex] and the ratio is [latex]r=\frac{2}{3}[/latex].

This means we can rewrite the series in standard form:

[latex]\displaystyle\sum _{n=1}^{\infty }\frac{4}{9}\cdot {\left(\frac{2}{3}\right)}^{n - 1}[/latex].

Since [latex]|r|=\frac{2}{3}<1[/latex], the series converges. Using our convergence formula:

[latex]\displaystyle\sum _{n=1}^{\infty }\frac{4}{9}\cdot {\left(\frac{2}{3}\right)}^{n - 1}=\frac{\frac{4}{9}}{1 - \frac{2}{3}}=\frac{\frac{4}{9}}{\frac{1}{3}}=\frac{4}{9}\cdot\frac{3}{1}=\frac{4}{3}[/latex]

Determine whether each of the following geometric series converges or diverges, and if it converges, find its sum.

  1. [latex]\displaystyle\sum _{n=1}^{\infty }\frac{{\left(-3\right)}^{n+1}}{{4}^{n - 1}}[/latex]
  2. [latex]\displaystyle\sum _{n=1}^{\infty }{e}^{2n}[/latex]

We now turn our attention to a nice application of geometric series. We show how they can be used to write repeating decimals as fractions of integers.

Use a geometric series to write [latex]3.\overline{26}[/latex] as a fraction of integers.