Different Types of Series

Learning Outcomes

  • Calculate the sum of a geometric series

The Harmonic Series

A useful series to know about is the harmonic series. The harmonic series is defined as

[latex]\displaystyle\sum _{n=1}^{\infty }\frac{1}{n}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots [/latex].

 

This series is interesting because it diverges, but it diverges very slowly. By this we mean that the terms in the sequence of partial sums [latex]\left\{{S}_{k}\right\}[/latex] approach infinity, but do so very slowly. We will show that the series diverges, but first we illustrate the slow growth of the terms in the sequence [latex]\left\{{S}_{k}\right\}[/latex] in the following table.

[latex]k[/latex] [latex]10[/latex] [latex]100[/latex] [latex]1000[/latex] [latex]10,000[/latex] [latex]100,000[/latex] [latex]1,000,000[/latex]
[latex]{S}_{k}[/latex] [latex]2.92897[/latex] [latex]5.18738[/latex] [latex]7.48547[/latex] [latex]9.78761[/latex] [latex]12.09015[/latex] [latex]14.39273[/latex]

Even after [latex]1,000,000[/latex] terms, the partial sum is still relatively small. From this table, it is not clear that this series actually diverges. However, we can show analytically that the sequence of partial sums diverges, and therefore the series diverges.

To show that the sequence of partial sums diverges, we show that the sequence of partial sums is unbounded. We begin by writing the first several partial sums:

[latex]\begin{array}{l}{S}_{1}=1\hfill \\ {S}_{2}=1+\frac{1}{2}\hfill \\ {S}_{3}=1+\frac{1}{2}+\frac{1}{3}\hfill \\ {S}_{4}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}.\hfill \end{array}[/latex]

 

Notice that for the last two terms in [latex]{S}_{4}[/latex],

[latex]\frac{1}{3}+\frac{1}{4}>\frac{1}{4}+\frac{1}{4}[/latex].

 

Therefore, we conclude that

[latex]{S}_{4}>1+\frac{1}{2}+\left(\frac{1}{4}+\frac{1}{4}\right)=1+\frac{1}{2}+\frac{1}{2}=1+2\left(\frac{1}{2}\right)[/latex].

 

Using the same idea for [latex]{S}_{8}[/latex], we see that

[latex]\begin{array}{cc}\hfill {S}_{8}& =1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}>1+\frac{1}{2}+\left(\frac{1}{4}+\frac{1}{4}\right)+\left(\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}\right)\hfill \\ & =1+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=1+3\left(\frac{1}{2}\right).\hfill \end{array}[/latex]

 

From this pattern, we see that [latex]{S}_{1}=1[/latex], [latex]{S}_{2}=1+\frac{1}{2}[/latex], [latex]{S}_{4}>1+2\left(\frac{1}{2}\right)[/latex], and [latex]{S}_{8}>1+3\left(\frac{1}{2}\right)[/latex]. More generally, it can be shown that [latex]{S}_{{2}^{j}}>1+j\left(\frac{1}{2}\right)[/latex] for all [latex]j>1[/latex]. Since [latex]1+j\left(\frac{1}{2}\right)\to \infty [/latex], we conclude that the sequence [latex]\left\{{S}_{k}\right\}[/latex] is unbounded and therefore diverges. In the previous section, we stated that convergent sequences are bounded. Consequently, since [latex]\left\{{S}_{k}\right\}[/latex] is unbounded, it diverges. Thus, the harmonic series diverges.

Algebraic Properties of Convergent Series

Since the sum of a convergent infinite series is defined as a limit of a sequence, the algebraic properties for series listed below follow directly from the algebraic properties for sequences.

Theorem: Algebraic Properties of Convergent Series

Let [latex]\displaystyle\sum _{n=1}^{\infty }{a}_{n}[/latex] and [latex]\displaystyle\sum _{n=1}^{\infty }{b}_{n}[/latex] be convergent series. Then the following algebraic properties hold.

  1. The series [latex]\displaystyle\sum _{n=1}^{\infty }\left({a}_{n}+{b}_{n}\right)[/latex] converges and [latex]\displaystyle\sum _{n=1}^{\infty }\left({a}_{n}+{b}_{n}\right)=\displaystyle\sum _{n=1}^{\infty }{a}_{n}+\displaystyle\sum _{n=1}^{\infty }{b}_{n}[/latex]. (Sum Rule)
  2. The series [latex]\displaystyle\sum _{n=1}^{\infty }\left({a}_{n}-{b}_{n}\right)[/latex] converges and [latex]\displaystyle\sum _{n=1}^{\infty }\left({a}_{n}-{b}_{n}\right)=\displaystyle\sum _{n=1}^{\infty }{a}_{n}-\displaystyle\sum _{n=1}^{\infty }{b}_{n}[/latex]. (Difference Rule)
  3. For any real number [latex]c[/latex], the series [latex]\displaystyle\sum _{n=1}^{\infty }c{a}_{n}[/latex] converges and [latex]\displaystyle\sum _{n=1}^{\infty }c{a}_{n}=c\displaystyle\sum _{n=1}^{\infty }{a}_{n}[/latex]. (Constant Multiple Rule)

Example: Using Algebraic Properties of Convergent Series

Evaluate

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

 

try it

Evaluate [latex]\displaystyle\sum _{n=1}^{\infty }\frac{5}{{2}^{n - 1}}[/latex].

Watch the following video to see the worked solution to the above Try IT.

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.

You can view the transcript for this segmented clip of “5.2.1” here (opens in new window).

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 r, the number r is called the ratio. We refer to a as the initial term because it is the first term in the series. For 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 ratio [latex]r=\frac{1}{2}[/latex].

In general, when does a geometric series converge? Consider the geometric series

[latex]\displaystyle\sum _{n=1}^{\infty }a{r}^{n - 1}[/latex]

 

when [latex]a>0[/latex]. Its sequence of partial sums [latex]\left\{{S}_{k}\right\}[/latex] is given by

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

 

Consider the case when [latex]r=1[/latex]. In that case,

[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 know [latex]ak\to \infty [/latex] as [latex]k\to \infty [/latex]. Therefore, the sequence of partial sums is unbounded and thus diverges. Consequently, the infinite series diverges for [latex]r=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]. Doing so, we see that

[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]

 

All the other terms cancel out.

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]. Therefore, for [latex]|r|<1[/latex], [latex]{S}_{k}\to \frac{a}{\left(1-r\right)}[/latex] and we have

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

 

If [latex]|r|\ge 1[/latex], [latex]{S}_{k}[/latex] diverges, and therefore

[latex]\displaystyle\sum _{n=1}^{\infty }a{r}^{n - 1}\text{ diverges if }|r|\ge 1[/latex].

 

Definition

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.

Geometric series sometimes appear in slightly different forms. For example, sometimes the index begins at a value other than [latex]n=1[/latex] or the exponent involves a linear expression for [latex]n[/latex] other than [latex]n - 1[/latex]. As long as we can rewrite the series in the form given by the harmonic series general equation, it is a geometric series. For example, consider the series

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

 

To see that this is a geometric series, we 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]. Therefore, the series can be written as

[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], this series converges, and its sum is given by

[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{4}{3}[/latex].

 

Example: Determining Convergence or Divergence of a Geometric Series

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]

try it

Determine whether the series [latex]\displaystyle\sum _{n=1}^{\infty }{\left(\frac{-2}{5}\right)}^{n - 1}[/latex] converges or diverges. If it converges, find its sum.

Watch the following video to see the worked solution to the above Try IT.

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.

You can view the transcript for this segmented clip of “5.2.2” here (opens in new window).

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.

Example: Writing Repeating Decimals as Fractions of Integers

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

try it

Write [latex]5.2\overline{7}[/latex] as a fraction of integers.