# 9.2: Infinite Series

- Page ID
- 2563

\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

We have seen that a sequence is an ordered set of terms. If you add these terms together, you get a series. In this section we define an infinite series and show how series are related to sequences. We also define what it means for a series to converge or diverge. We introduce one of the most important types of series: the geometric series. We will use geometric series in the next chapter to write certain functions as polynomials with an infinite number of terms. This process is important because it allows us to evaluate, differentiate, and integrate complicated functions by using polynomials that are easier to handle. We also discuss the harmonic series, arguably the most interesting divergent series because it just fails to converge.

## Sums and Series

An infinite series is a sum of infinitely many terms and is written in the form

\(\displaystyle \sum_{n=1}^∞a_n=a_1+a_2+a_3+⋯.\)

But what does this mean? We cannot add an infinite number of terms in the same way we can add a finite number of terms. Instead, the value of an infinite series is defined in terms of the limit of partial sums. A partial sum of an infinite series is a finite sum of the form

\(\displaystyle \sum_{n=1}^ka_n=a_1+a_2+a_3+⋯+a_k.\)

To see how we use partial sums to evaluate infinite series, consider the following example. Suppose oil is seeping into a lake such that \(\displaystyle 1000\) gallons enters the lake the first week. During the second week, an additional \(\displaystyle 500\) gallons of oil enters the lake. The third week, \(\displaystyle 250\) more gallons enters the lake. Assume this pattern continues such that each week half as much oil enters the lake as did the previous week. If this continues forever, what can we say about the amount of oil in the lake? Will the amount of oil continue to get arbitrarily large, or is it possible that it approaches some finite amount? To answer this question, we look at the amount of oil in the lake after \(\displaystyle k\) weeks. Letting \(\displaystyle S_k\) denote the amount of oil in the lake (measured in thousands of gallons) after \(\displaystyle k\) weeks, we see that

\(\displaystyle S_1=1\)

\(\displaystyle S_2=1+0.5=1+\frac{1}{2}\)

\(\displaystyle S_3=1+0.5+0.25=1+\frac{1}{2}+\frac{1}{4}\)

\(\displaystyle S_4=1+0.5+0.25+0.125=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}\)

\(\displaystyle S_5=1+0.5+0.25+0.125+0.0625=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}.\)

Looking at this pattern, we see that the amount of oil in the lake (in thousands of gallons) after \(\displaystyle k\) weeks is

\(\displaystyle S_k=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+⋯+\frac{1}{2^{k−1}}=\sum_{n=1}^k(\frac{1}{2})^{n−1}.\)

We are interested in what happens as \(\displaystyle k→∞.\) Symbolically, the amount of oil in the lake as \(\displaystyle k→∞\) is given by the infinite series

\(\displaystyle \sum_{n=1}^∞(\frac{1}{2})^{n−1}=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+⋯.\)

At the same time, as \(\displaystyle k→∞\), the amount of oil in the lake can be calculated by evaluating \(\displaystyle \lim_{k→∞}S_k\). Therefore, the behavior of the infinite series can be determined by looking at the behavior of the sequence of partial sums \(\displaystyle {S_k}\). If the sequence of partial sums \(\displaystyle {S_k}\) converges, we say that the infinite series converges, and its sum is given by \(\displaystyle \lim_{k→∞}S_k\). If the sequence \(\displaystyle {S_k}\) diverges, we say the infinite series diverges. We now turn our attention to determining the limit of this sequence \(\displaystyle {S_k}\).

First, simplifying some of these partial sums, we see that

\(\displaystyle S_1=1\)

\(\displaystyle S_2=1+\frac{1}{2}=\frac{3}{2}\)

\(\displaystyle S_3=1+\frac{1}{2}+\frac{1}{4}=\frac{7}{4}\)

\(\displaystyle S_4=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}=\frac{15}{8}\)

\(\displaystyle S_5=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}=\frac{31}{16}.\)

Plotting some of these values in Figure, it appears that the sequence \(\displaystyle {S_k}\) could be approaching 2.

**Figure \(\PageIndex{1}\):** The graph shows the sequence of partial sums \(\displaystyle {S_k}\). It appears that the sequence is approaching the value \(\displaystyle 2\).

Let’s look for more convincing evidence. In the following table, we list the values of \(\displaystyle S_k\) for several values of \(\displaystyle k\).

\(\displaystyle k\) | 5 | 10 | 15 | 20 |

\(\displaystyle S_k\) | 1.9375 | 1.998 | 1.999939 | 1.999998 |

These data supply more evidence suggesting that the sequence \(\displaystyle {S_k}\) converges to \(\displaystyle 2\). Later we will provide an analytic argument that can be used to prove that \(\displaystyle \lim_{k→∞}S_k=2\). For now, we rely on the numerical and graphical data to convince ourselves that the sequence of partial sums does actually converge to \(\displaystyle 2\). Since this sequence of partial sums converges to \(\displaystyle 2\), we say the infinite series converges to \(\displaystyle 2\) and write

\(\displaystyle \sum_{n=1}^∞(\frac{1}{2})^{n−1}=2.\)

Returning to the question about the oil in the lake, since this infinite series converges to \(\displaystyle 2\), we conclude that the amount of oil in the lake will get arbitrarily close to \(\displaystyle 2000\) gallons as the amount of time gets sufficiently large.

This series is an example of a geometric series. We discuss geometric series in more detail later in this section. First, we summarize what it means for an infinite series to converge.

Definition

An **infinite series** is an expression of the form

\[\sum_{n=1}^∞a_n=a_1+a_2+a_3+⋯.\]

For each positive integer \(\displaystyle k\), the sum

\[\displaystyle S_k=\sum_{n=1}^ka_n=a_1+a_2+a_3+⋯+a_k\]

is called the \(\displaystyle kth\) **partial sum **of the infinite series. The partial sums form a sequence \(\displaystyle {S_k}\). If the sequence of partial sums converges to a real number \(\displaystyle S\), the infinite series converges. If we can describe the **convergence of a series** to \(\displaystyle S\), we call \(\displaystyle S\) the sum of the series, and we write

\[\displaystyle \sum_{n=1}^∞a_n=S.\]

If the sequence of partial sums diverges, we have the** divergence of a series.**

This website shows a more whimsical approach to series.

Note that the index for a series need not begin with \(\displaystyle n=1\) but can begin with any value. For example, the series

\[\displaystyle \sum_{n=1}∞^(\frac{1}{2})^{n−1}\]

can also be written as

\(\displaystyle \sum_{n=0}^∞(\frac{1}{2})^n\) or \(\displaystyle \sum_{n=5}^∞(\frac{1}{2})^{n−5}.\)

Often it is convenient for the index to begin at \(\displaystyle 1\), so if for some reason it begins at a different value, we can reindex by making a change of variables. For example, consider the series

\(\displaystyle \sum_{n=2}^∞\frac{1}{n^2}\).

By introducing the variable \(\displaystyle m=n−1\), so that \(\displaystyle n=m+1,\) we can rewrite the series as

\(\displaystyle \sum_{m=1}^∞\frac{1}{(m+1)^2}\).

Example \(\displaystyle \PageIndex{1}\): Evaluating Limits of Sequences of Partial Sums

For each of the following series, use the sequence of partial sums to determine whether the series converges or diverges.

a. \(\displaystyle \sum_{n=1}^∞\frac{n}{n+1}\)

b \(\displaystyle \sum_{n=1}^∞(−1)^n\)

c \(\displaystyle \sum_{n=1}^∞\frac{1}{n(n+1)}\)

**Solution**

a. The sequence of partial sums \(\displaystyle {S_k}\) satisfies

\(\displaystyle S_1=\frac{1}{2}\)

\(\displaystyle S_2=\frac{1}{2}+\frac{2}{3}\)

\(\displaystyle S_3=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}\)

\(\displaystyle S_4=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+\frac{4}{5}\).

Notice that each term added is greater than \(\displaystyle 1/2\). As a result, we see that

\(\displaystyle S_1=\frac{1}{2}\)

\(\displaystyle S_2=\frac{1}{2}+\frac{2}{3}>\frac{1}{2}+\frac{1}{2}=2(\frac{1}{2})\)

\(\displaystyle S_3=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}>\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=3(\frac{1}{2})\)

\(\displaystyle S_4=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+\frac{4}{5}>\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=4(\frac{1}{2}).\)

From this pattern we can see that \(\displaystyle S_k>k(\frac{1}{2})\) for every integer \(\displaystyle k\). Therefore, \(\displaystyle {S_k}\) is unbounded and consequently, diverges. Therefore, the infinite series \(\displaystyle \sum^∞_{n=1}n/(n+1)\) diverges.

b. The sequence of partial sums \(\displaystyle {S_k}\) satisfies

\(\displaystyle S_1=−1\)

\(\displaystyle S_2=−1+1=0\)

\(\displaystyle S_3=−1+1−1=−1\)

\(\displaystyle S_4=−1+1−1+1=0.\)

From this pattern we can see the sequence of partial sums is

\(\displaystyle {S_k}={−1,0,−1,0,…}.\)

Since this sequence diverges, the infinite series \(\displaystyle \sum^∞_{n=1}(−1)^n\) diverges.

c. The sequence of partial sums \(\displaystyle {S_k}\) satisfies

\(\displaystyle S_1=\frac{1}{1⋅2}=\frac{1}{2}\)

\(\displaystyle S_2=\frac{1}{1⋅2}+\frac{1}{2⋅3}=\frac{1}{2}+\frac{1}{6}=\frac{2}{3}\)

\(\displaystyle S_3=\frac{1}{1⋅2}+\frac{1}{2⋅3}+\frac{1}{3⋅4}=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}=\frac{3}{4}\)

\(\displaystyle S_4=\frac{1}{1⋅2}+\frac{1}{2⋅3}+\frac{1}{3⋅4}+\frac{1}{4⋅5}=\frac{4}{5}\)

\(\displaystyle S_5=\frac{1}{1⋅2}+\frac{1}{2⋅3}+\frac{1}{3⋅4}+\frac{1}{4⋅5}+\frac{1}{5⋅6}=\frac{5}{6}.\)

From this pattern, we can see that the \(\displaystyle kth\) partial sum is given by the explicit formula

\(\displaystyle S_k=\frac{k}{k+1}\).

Since \(\displaystyle k/(k+1)→1,\) we conclude that the sequence of partial sums converges, and therefore the infinite series converges to \(\displaystyle 1\). We have

\(\displaystyle \sum_{n=1}^∞\frac{1}{n(n+1)}=1.\)

Exercise \(\PageIndex{1}\)

Determine whether the series \(\displaystyle \sum^∞_{n=1}(n+1)/n\) converges or diverges.

**Hint**-
Look at the sequence of partial sums.

**Answer**-
The series diverges because the \(\displaystyle kth\) partial sum \(\displaystyle S_k>k\).

## The Harmonic Series

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

\(\displaystyle \sum_{n=1}^∞\frac{1}{n}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+⋯.\)

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 \(\displaystyle {S_k}\) 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 \(\displaystyle {S_k}\) in the following table.

\(\displaystyle k\) | 10 | 100 | 1000 | 10,00 | 100,000 | 1,000,000 |

\(\displaystyle S_k\) | 2.92897 | 5.18738 | 7.48547 | 9.78761 | 12.09015 | 14.39273 |

Even after \(\displaystyle 1,000,000\) 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:

\(\displaystyle S_1=1\)

\(\displaystyle S_2=1+\frac{1}{2}\)

\(\displaystyle S_3=1+\frac{1}{2}+\frac{1}{3}\)

\(\displaystyle S_4=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\).

Notice that for the last two terms in \(\displaystyle S_4\),

\(\displaystyle \frac{1}{3}+\frac{1}{4}>\frac{1}{4}+\frac{1}{4}\).

Therefore, we conclude that

\(\displaystyle S_4>1+\frac{1}{2}+(\frac{1}{4}+\frac{1}{4})=1+\frac{1}{2}+\frac{1}{2}=1+2(\frac{1}{2})\).

Using the same idea for \(\displaystyle S_8\), we see that

\(\displaystyle S_8=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}>1+\frac{1}{2}+(\frac{1}{4}+\frac{1}{4})+(\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8})=1+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=1+3(\frac{1}{2})\).

From this pattern, we see that \(\displaystyle S_1=1, S_2=1+1/2, S_4>1+2(1/2),\) and \(\displaystyle S_8>1+3(1/2)\). More generally, it can be shown that \(\displaystyle S_{2^j}>1+j(1/2)\) for all \(\displaystyle j>1\). Since \(\displaystyle 1+j(1/2)→∞,\) we conclude that the sequence \(\displaystyle {S_k}\) is unbounded and therefore diverges. In the previous section, we stated that convergent sequences are bounded. Consequently, since \(\displaystyle {S_k}\) 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.

Algebraic Properties of Convergent Series

Let \(\displaystyle \sum_{n=1}^∞(a_n

i. The series \(\displaystyle \sum_{n=1}^∞(a_n+b_n)\) converges and \(\displaystyle \sum^∞_{n=1}(a_n+b_n)=\sum^∞_{n=1}a_n+\sum^∞_{n=1}b_n.\) (Sum Rule)

ii. The series \(\displaystyle \sum_{n=1}^∞(a_n−b_n)\) converges and \(\displaystyle \sum^∞_{n=1}(a_n−b_n)=\sum^∞_{n=1}a_n−\sum^∞_{n=1}b_n.\) (Difference Rule)

iii. For any real number \(\displaystyle c\), the series \(\displaystyle \sum_{n=1}^∞ca_n\) converges and (\sum^∞_{n=1}ca_n=c\sum^∞_{n=1}a_n\). (Constant Multiple Rule)

Example \(\displaystyle \PageIndex{2}\):Using Algebraic Properties of Convergent Series

**Evaluate**

\(\displaystyle \sum_{n=1}^∞[\frac{3}{n(n+1)}+(\frac{1}{2})^{n−2}].\)

**Solution**

We showed earlier that

\(\displaystyle \sum_{n=1}^∞\frac{1}{n(n+1)}\)

and

\(\displaystyle \sum_{n=1}^∞(\frac{1}{2})^{n−1}=2.\)

Since both of those series converge, we can apply the properties of Note to evaluate

\(\displaystyle \sum_{n=1}^∞[\frac{3}{n(n+1)}+(\frac{1}{2})^{n−2}].\)

Using the sum rule, write

\(\displaystyle \sum_{n=1}^∞[\frac{3}{n(n+1)}+(\frac{1}{2})^{n−2}]=\sum_{n=1}^∞\frac{3}{n(n+1)}+\sum_{n=1}^∞(\frac{1}{2})^{n−2}.\)

Then, using the constant multiple rule and the sums above, we can conclude that

\(\displaystyle \sum^∞_{n=1}\frac{3}{n(n+1)}+\sum^∞_{n=1}(\frac{1}{2})^{n−2}=3\sum^∞_{n=1}\frac{1}{n(n+1)}+(\frac{1}{2})^{−1}\sum^∞_{n=1}(\frac{1}{2})^{n−1}=3(1)+(\frac{1}{2})^{−1}(2)=3+2(2)=7.\)

Exercise \(\PageIndex{2}\)

Evaluate \(\displaystyle \sum^∞_{n=1}\frac{5}{2^{n−1}}\).

**Hint**-
Rewrite as \(\displaystyle \sum^∞_{n=1}5(\frac{1}{2})^{n−1}\).

**Answer**-
10.

## Geometric Series

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

\(\displaystyle a+ar+ar^2+ar^3+⋯=\sum_{n=1}^∞ar^{n−1}.\)

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

\(\displaystyle \sum_{n=1}^∞(\frac{1}{2})^{n−1}=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+⋯\)

is a geometric series with initial term \(\displaystyle a=1\) and ratio \(\displaystyle r=1/2\).

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

\(\displaystyle \sum_{n=1}^∞ar^{n−1}\)

when \(\displaystyle a>0\). Its sequence of partial sums \(\displaystyle {S_k}\) is given by

\(\displaystyle S_k=\sum_{n=1}^kar^{n−1}=a+ar+ar^2+⋯+ar^{k−1}.\)

Consider the case when \(\displaystyle r=1.\) In that case,

\(\displaystyle S_k=a+a(1)+a(1)^2+⋯+a(1)^{k−1}=ak.\)

Since \(\displaystyle a>0\), we know \(\displaystyle ak→∞\) as \(\displaystyle k→∞\). Therefore, the sequence of partial sums is unbounded and thus diverges. Consequently, the infinite series diverges for \(\displaystyle r=1\). For \(\displaystyle r≠1\), to find the limit of \(\displaystyle {S_k}\), multiply Equation by \(\displaystyle 1−r\). Doing so, we see that

\(\displaystyle (1−r)S_k=a(1−r)(1+r+r^2+r^3+⋯+r^{k−1})=a[(1+r+r^2+r^3+⋯+r^{k−1})−(r+r^2+r^3+⋯+r^k)]=a(1−r^k).\)

All the other terms cancel out.

Therefore,

\(\displaystyle S_k=\frac{a(1−r^k)}{1−r}\) for \(\displaystyle r≠1\).

From our discussion in the previous section, we know that the geometric sequence \(\displaystyle r^k→0\) if \(\displaystyle |r|<1\) and that \(\displaystyle r^k\) diverges if \(\displaystyle |r|>1\) or \(\displaystyle r=±1\). Therefore, for \(\displaystyle |r|<1, S_k→a/(1−r)\) and we have

\(\displaystyle \sum_{n=1}^∞ar^{n−1}=\frac{a}{1−r}\) if \(\displaystyle |r|<1.\)

If \(\displaystyle |r|≥1, S_k\) diverges, and therefore

\(\displaystyle \sum_{n=1}^∞ar^{n−1}\) diverges if \(\displaystyle |r|≥1.\)

Definition

A geometric series is a series of the form

\(\displaystyle \sum_{n=1}^∞ar^{n−1}=a+ar+ar^2+ar^3+⋯.\)

If \(\displaystyle |r|<1\), the series converges, and

\(\displaystyle \sum_{n=1}^∞ar^{n−1}=\frac{a}{1−r}\) for \(\displaystyle |r|<1\).

If \(\displaystyle |r|≥1\), the series diverges.

Geometric series sometimes appear in slightly different forms. For example, sometimes the index begins at a value other than \(\displaystyle n=1\) or the exponent involves a linear expression for \(\displaystyle n\) other than \(\displaystyle n−1\). As long as we can rewrite the series in the form given by Equation, it is a geometric series. For example, consider the series

\(\displaystyle \sum_{n=0}^∞(\frac{2}{3})^{n+2}.\)

To see that this is a geometric series, we write out the first several terms:

\(\displaystyle \sum_{n=0}^∞(\frac{2}{3})^{n+2}=(\frac{2}{3})^2+(\frac{2}{3})^3+(\frac{2}{3})^4+⋯=\frac{4}{9}+\frac{4}{9}⋅(\frac{2}{3})+\frac{4}{9}⋅(\frac{2}{3})^2+⋯.\)

We see that the initial term is \(\displaystyle a=4/9\) and the ratio is \(\displaystyle r=2/3.\) Therefore, the series can be written as

\(\displaystyle \sum_{n=1}^∞\frac{4}{9}⋅(\frac{2}{3})^{n−1}.\)

Since \(\displaystyle r=2/3<1\), this series converges, and its sum is given by

\(\displaystyle \sum_{n=1}^∞\frac{4}{9}⋅(\frac{2}{3})^{n−1}=\frac{4/9}{1−2/3}=\frac{4}{3}.\)

Example \(\displaystyle \PageIndex{3}\): 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.

a. \(\displaystyle \sum^∞_{n=1}\frac{(−3)^{n+1}}{4^{n−1}}\)

b. \(\displaystyle \sum^∞_{n=1}e^{2n}\)

**Solution**

a. Writing out the first several terms in the series, we have

\(\displaystyle \sum_{n=1}^∞\frac{(−3)^{n+1}}{4^{n−1}}=\frac{(−3)^2}{4^0}+\frac{(−3)^3}{4}+\frac{(−3)^4}{4^2}+⋯=(−3)^2+(−3)^2⋅(\frac{−3}{4})+(−3)^2⋅(\frac{−3}{4})^2+⋯=9+9⋅(\frac{−3}{4})+9⋅(\frac{−3}{4})^2+⋯.\)

The initial term \(\displaystyle a=−3\) and the ratio \(\displaystyle r=−3/4\). Since \(\displaystyle |r|=3/4<1\), the series converges to

\(\displaystyle \frac{9}{1−(−3/4)}=\frac{9}{7/4}=\frac{36}{7}\).

b. Writing this series as

\(\displaystyle e^2\sum_{n=1}^∞(e^2)^{n−1}\)

we can see that this is a geometric series where \(\displaystyle r=e^2>1.\) Therefore, the series diverges.

Exercise \(\PageIndex{3}\)

Determine whether the series \(\displaystyle \sum_{n=1}^∞(\frac{−2}{5})^{n−1}\) converges or diverges. If it converges, find its sum.

**Hint**-
\(\displaystyle r=−2/5\)

**Answer**-
\(\displaystyle 5/7\)

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 \(\displaystyle \PageIndex{4}\): Writing Repeating Decimals as Fractions of Integers

Use a geometric series to write \(\displaystyle 3.\bar{26}\) as a fraction of integers.

**Solution**

Since \(\displaystyle 3.\bar{26}—=3.262626…,\) first we write

\(\displaystyle 3.262626…=3+\frac{26}{100}+\frac{26}{1000}+\frac{26}{100,000}+⋯=3+\frac{26}{10^2}+\frac{26}{10^4}+\frac{26}{10^6}+⋯.\)

Ignoring the term 3, the rest of this expression is a geometric series with initial term \(\displaystyle a=26/10^2\) and ratio \(\displaystyle r=1/10^2.\) Therefore, the sum of this series is

\(\displaystyle \frac{26/10^2}{1−(1/10^2)}=\frac{26/10^2}{99/10^2}=\frac{26}{99}\).

Thus,

\(\displaystyle 3.262626…=3+\frac{26}{99}=\frac{323}{99}\).

Exercise \(\PageIndex{4}\)

Write \(\displaystyle 5.2\bar{7}\) as a fraction of integers.

**Hint**-
By expressing this number as a series, find a geometric series with initial term \(\displaystyle a=7/100\) and ratio \(\displaystyle r=1/10\).

**Answer**-
\(\displaystyle 475/90\)

Example \(\displaystyle \PageIndex{5}\): Chapter Opener: Finding the Area of the Koch Snowflake

Define a sequence of figures \(\displaystyle {F_n}\) recursively as follows (Figure). Let \(\displaystyle F_0\) be an equilateral triangle with sides of length \(\displaystyle 1\). For \(\displaystyle n≥1\), let \(\displaystyle F_n\) be the curve created by removing the middle third of each side of \(\displaystyle F_{n−1}\) and replacing it with an equilateral triangle pointing outward. The limiting figure as \(\displaystyle n→∞\) is known as **Koch’s snowflake.**

**Figure \(\PageIndex{2}\):** The first four figures, \(\displaystyle F_0,F_1,F_2\), and \(\displaystyle F_3\), in the construction of the Koch snowflake.

a. Find the length \(\displaystyle L_n\) of the perimeter of \(\displaystyle F_n\). Evaluate \(\displaystyle \lim_{n→∞}L_n\) to find the length of the perimeter of Koch’s snowflake.

b. Find the area \(\displaystyle A_n\) of figure \(\displaystyle F_n\). Evaluate \(\displaystyle \lim_{n→∞}A_n\) to find the area of Koch’s snowflake.

**Solution**

a. Let \(\displaystyle N_n\) denote the number of sides of figure \(\displaystyle F_n\). Since \(\displaystyle F_0\) is a triangle, \(\displaystyle N_0=3\). Let ln denote the length of each side of \(\displaystyle F_n\). Since \(\displaystyle F_0\) is an equilateral triangle with sides of length \(\displaystyle l_0=1\), we now need to determine \(\displaystyle N_1\) and \(\displaystyle l_1\). Since \(\displaystyle F_1\) is created by removing the middle third of each side and replacing that line segment with two line segments, for each side of \(\displaystyle F_0\), we get four sides in \(\displaystyle F_1\). Therefore, the number of sides for \(\displaystyle F_1\) is

\(\displaystyle N_1=4⋅3\).

Since the length of each of these new line segments is \(\displaystyle 1/3\) the length of the line segments in \(\displaystyle F_0\), the length of the line segments for \(\displaystyle F_1\) is given by

\(\displaystyle l_1=\frac{1}{3}⋅1=\frac{1}{3}\).

Similarly, for \(\displaystyle F_2\), since the middle third of each side of \(\displaystyle F_1\) is removed and replaced with two line segments, the number of sides in \(\displaystyle F_2\) is given by

\(\displaystyle N_2=4N_1=4(4⋅3)=4^2⋅3.\)

Since the length of each of these sides is \(\displaystyle 1/3\) the length of the sides of \(\displaystyle F_1\), the length of each side of figure \(\displaystyle F_2\) is given by

\(\displaystyle l_2=\frac{1}{3}⋅l1=\frac{1}{3}⋅\frac{1}{3}=(\frac{1}{3})^2\).

More generally, since \(\displaystyle F_n\) is created by removing the middle third of each side of \(\displaystyle F_{n−1}\) and replacing that line segment with two line segments of length \(\displaystyle \frac{1}{3}l_{n−1}\) in the shape of an equilateral triangle, we know that \(\displaystyle N_n=4N_{n−1}\) and \(\displaystyle l_n=\frac{l_{n−1}}{3}\). Therefore, the number of sides of figure \(\displaystyle F_n\) is

\(\displaystyle N_n=4^n⋅3\)

and the length of each side is

\(\displaystyle l_n=(\frac{1}{3})^n\).

Therefore, to calculate the perimeter of \(\displaystyle F_n\), we multiply the number of sides \(\displaystyle N_n\) and the length of each side \(\displaystyle l_n\). We conclude that the perimeter of \(\displaystyle F_n\) is given by

\(\displaystyle L_n=N_n⋅l_n=3⋅(\frac{4}{3})^n\).

Therefore, the length of the perimeter of Koch’s snowflake is

\(\displaystyle L=\lim_{n→∞}L_n=∞.\)

b. Let \(\displaystyle T_n\) denote the area of each new triangle created when forming \(\displaystyle F_n\). For \(\displaystyle n=0, T_0\) is the area of the original equilateral triangle. Therefore, \(\displaystyle T_0=A_0=\sqrt{3}/4\). For \(\displaystyle n≥1\), since the lengths of the sides of the new triangle are \(\displaystyle 1/3\) the length of the sides of \(\displaystyle F_{n−1}\), we have

\(\displaystyle T_n=(\frac{1}{3})^2T_{n−1}=\frac{1}{9}⋅T_{n−1}.\)

Therefore, \(\displaystyle T_n=(\frac{1}{9})^n⋅\frac{\sqrt{3}}{4}\). Since a new triangle is formed on each side of \(\displaystyle F_{n−1}\),

\(\displaystyle A_n=A_{n−1}+N_{n−1}⋅T_n=A_{n−1}+(3⋅4_{n−1})⋅(\frac{1}{9})^n⋅\frac{\sqrt{3}}{4}=A_{n−1}+\frac{3}{4}⋅(\frac{4}{9})^n⋅\frac{\sqrt{3}}{4}\).

Writing out the first few terms \(\displaystyle A_0,A_1,A_2,\) we see that

\(\displaystyle A_0=\frac{\sqrt{3}}{4}\)

\(\displaystyle A_1=A_0+\frac{3}{4}⋅(\frac{4}{9})⋅\frac{\sqrt{3}}{4}=\frac{\sqrt{3}}{4}+\frac{3}{4}⋅(\frac{4}{9})⋅\frac{\sqrt{3}}{4}=\frac{\sqrt{3}}{4}[1+\frac{3}{4}⋅(\frac{4}{9})]\)

\(\displaystyle A_2=A_1+\frac{3}{4}⋅(\frac{4}{9})^2⋅\frac{\sqrt{3}}{4}=\frac{\sqrt{3}}{4}[1+\frac{3}{4}⋅(\frac{4}{9})]+\frac{3}{4}⋅(\frac{4}{9})^2⋅\frac{\sqrt{3}}{4}=\frac{\sqrt{3}}{4}[1+\frac{3}{4}⋅(\frac{4}{9})+\frac{3}{4}⋅(\frac{4}{9})^2]\).

More generally,

\(\displaystyle A)n=\frac{\sqrt{3}}{4}[1+\frac{3}{4}(\frac{4}{9}+(\frac{4}{9})^2+⋯+(\frac{4}{9})^n)]\).

Factoring \(\displaystyle 4/9\) out of each term inside the inner parentheses, we rewrite our expression as

\(\displaystyle A_n=\frac{\sqrt{3}}{4}[1+\frac{1}{3}(1+\frac{4}{9}+(\frac{4}{9})^2+⋯+(\frac{4}{9})^{n−1})]\).

The expression \(\displaystyle 1+(\frac{4}{9})+(\frac{4}{9})^2+⋯+(\frac{4}{9})^{n−1}\) is a geometric sum. As shown earlier, this sum satisfies

\(\displaystyle 1+\frac{4}{9}+(\frac{4}{9})^2+⋯+(\frac{4}{9})^{n−1}=\frac{1−(4/9)^n}{1−(4/9)}.\)

Substituting this expression into the expression above and simplifying, we conclude that

\(\displaystyle A_n=\frac{\sqrt{3}}{4}[1+\frac{1}{3}(\frac{1−(4/9)^n}{1−(4/9)})]=\frac{\sqrt{3}}{4}[\frac{8}{5}−\frac{3}{5}(\frac{4}{9})^n].\)

Therefore, the area of Koch’s snowflake is

\(\displaystyle A=\lim_{n→∞}A_n=\frac{2\sqrt{3}}{5}.\)

**Analysis**

The Koch snowflake is interesting because it has finite area, yet infinite perimeter. Although at first this may seem impossible, recall that you have seen similar examples earlier in the text. For example, consider the region bounded by the curve \(\displaystyle y=1/x^2\) and the \(\displaystyle x\)-axis on the interval \(\displaystyle [1,∞).\) Since the improper integral

\(\displaystyle ∫^∞_1\frac{1}{x^2}dx\)

converges, the area of this region is finite, even though the perimeter is infinite.

## Telescoping Series

Consider the series \(\displaystyle \sum_{n=1}^∞\frac{1}{n(n+1)}.\) We discussed this series in Example, showing that the series converges by writing out the first several partial sums \(\displaystyle S_1,S_2,…,S_6\) and noticing that they are all of the form \(\displaystyle S_k=\frac{k}{k+1}\). Here we use a different technique to show that this series converges. By using partial fractions, we can write

\(\displaystyle \frac{1}{n(n+1)}=\frac{1}{n}−\frac{1}{n+1}\).

Therefore, the series can be written as

\(\displaystyle \sum_{n=1}^∞[\frac{1}{n}−\frac{1}{n+1}]=(1+\frac{1}{2})+(\frac{1}{2}−\frac{1}{3})+(\frac{1}{3}−\frac{1}{4})+⋯.\)

Writing out the first several terms in the sequence of partial sums \(\displaystyle {S_k},\) we see that

\(\displaystyle S_1=1−\frac{1}{2}\)

\(\displaystyle S_2=(1−\frac{1}{2})+(\frac{1}{2}−\frac{1}{3})=1−\frac{1}{3}\)

\(\displaystyle S_3=(1−\frac{1}{2})+(\frac{1}{2}−\frac{1}{3})+(\frac{1}{3}−\frac{1}{4})=1−\frac{1}{4}\).

In general,

\(\displaystyle S_k=(1−\frac{1}{2})+(\frac{1}{2}−\frac{1}{3})+(\frac{1}{3}−\frac{1}{4})+⋯+(\frac{1}{k}−\frac{1}{k+1})=1−\frac{1}{k+1}\).

We notice that the middle terms cancel each other out, leaving only the first and last terms. In a sense, the series collapses like a spyglass with tubes that disappear into each other to shorten the telescope. For this reason, we call a series that has this property a telescoping series. For this series, since \(\displaystyle S_k=1−1/(k+1)\) and \(\displaystyle 1/(k+1)→0\) as \(\displaystyle k→∞\), the sequence of partial sums converges to \(\displaystyle 1\), and therefore the series converges to \(\displaystyle 1\).

Definition

A **telescoping series** is a series in which most of the terms cancel in each of the partial sums, leaving only some of the first terms and some of the last terms.

For example, any series of the form

\(\displaystyle \sum_{n=1}^∞[b_n−b_{n+1}]=(b_1−b_2)+(b_2−b_3)+(b_3−b_4)+⋯\)

is a telescoping series. We can see this by writing out some of the partial sums. In particular, we see that

\(\displaystyle S_1=b_1−b_2\)

\(\displaystyle S_2=(b_1−b_2)+(b_2−b_3)=b_1−b_3\)

\(\displaystyle S_3=(b_1−b_2)+(b_2−b_3)+(b_3−b_4)=b_1−b_4.\)

In general, the* kth *partial sum of this series is

\(\displaystyle S_k=b_1−b_{k+1}\).

Since the *kth* partial sum can be simplified to the difference of these two terms, the sequence of partial sums \(\displaystyle {S_k}\) will converge if and only if the sequence \(\displaystyle {b_{k+1}}\) converges. Moreover, if the sequence \(\displaystyle b_{k+1}\) converges to some finite number B, then the sequence of partial sums converges to \(\displaystyle b_1−B\), and therefore

\(\displaystyle \sum_{n=1}^∞[b_n−b_{n+1}]=b_1−B.\)

In the next example, we show how to use these ideas to analyze a telescoping series of this form.

Example \(\displaystyle \PageIndex{6}\): Evaluating a Telescoping Series

Determine whether the telescoping series

\(\displaystyle \sum_{n=1}^∞[cos(\frac{1}{n})−cos(\frac{1}{n+1})]\)

converges or diverges. If it converges, find its sum.

**Solution**

By writing out terms in the sequence of partial sums, we can see that

\(\displaystyle S_1=cos(1)−cos(\frac{1}{2})\)

\(\displaystyle S_2=(cos(1)−cos(\frac{1}{2}))+(cos(\frac{1}{2})−cos(\frac{1}{3}))=cos(1)−cos(\frac{1}{3})\)

\(\displaystyle S_3=(cos(1)−cos(\frac{1}{2}))+(cos(\frac{1}{2})−cos(\frac{1}{3}))+(cos(\frac{1}{3})−cos(\frac{1}{4}))\)

\(\displaystyle =cos(1)−cos(\frac{1}{4})\).

In general,

\(\displaystyle S_k=cos(1)−cos(\frac{1}{k+1})\).

Since \(\displaystyle 1/(k+1)→0\) as \(\displaystyle k→∞\) and \(\displaystyle cosx\) is a continuous function, \(\displaystyle cos(1/(k+1))→cos(0)=1\). Therefore, we conclude that \(\displaystyle S_k→cos(1)−1\). The telescoping series converges and the sum is given by

\(\displaystyle \sum_{n=1}^∞[cos(\frac{1}{n})−cos(\frac{1}{n+1})]=cos(1)−1.\)

Exercise \(\PageIndex{5}\)

Determine whether \(\displaystyle \sum^∞_{n=1}[e^{1/n}−e^{1/(n+1)}]\) converges or diverges. If it converges, find its sum.

**Hint**-
Write out the sequence of partial sums to see which terms cancel.

**Answer**-
\(\displaystyle e−1\)

Euler’s Constant

We have shown that the harmonic series \(\displaystyle \sum^∞_{n=1}\frac{1}{n}\) diverges. Here we investigate the behavior of the partial sums \(\displaystyle S_k\) as \(\displaystyle k→∞.\) In particular, we show that they behave like the natural logarithm function by showing that there exists a constant \(\displaystyle γ\) such that

\(\displaystyle \sum_{n=1}^k\frac{1}{n}−lnk→γ\) as \(\displaystyle k→∞.\)

This constant \(\displaystyle γ\) is known as **Euler’s constant.**

1. Let \(\displaystyle T_k=\sum_{n=1}^k\frac{1}{n}−lnk.\) Evaluate \(\displaystyle T_k\) for various values of \(\displaystyle k\).

2. For \(\displaystyle T_k\) as defined in part 1. show that the sequence \(\displaystyle {T_k}\) converges by using the following steps.

a. Show that the sequence \(\displaystyle {T_k}\) is monotone decreasing. (Hint: Show that \(\displaystyle ln(1+1/k>1/(k+1))\)

b. Show that the sequence \(\displaystyle {T_k}\) is bounded below by zero. (Hint: Express \(\displaystyle lnk\) as a definite integral.)

c. Use the Monotone Convergence Theorem to conclude that the sequence \(\displaystyle {T_k}\) converges. The limit \(\displaystyle γ\) is Euler’s constant.

3. Now estimate how far \(\displaystyle T_k\) is from \(\displaystyle γ\) for a given integer \(\displaystyle k\). Prove that for \(\displaystyle k≥1, 0<T_k−γ≤1/k\) by using the following steps.

a. Show that \(\displaystyle ln(k+1)−lnk<1/k.\)

b. Use the result from part a. to show that for any integer \(\displaystyle k\),

\(\displaystyle T_k−T_{k+1}<\frac{1}{k}−\frac{1}{k+1}.\)

c. For any integers \(\displaystyle k\) and \(\displaystyle j\) such that \(\displaystyle j>k\), express \(\displaystyle T_k−T_j\) as a telescoping sum by writing

\(\displaystyle T_k−T_j=(T_k−T_{k+1})+(T_{k+1}−T_{k+2})+(T_{k+2}−T_{k+3})+⋯+(T_{j−1}−T_j).\)

Use the result from part b. combined with this telescoping sum to conclude that

\(\displaystyle T_k−T_j<\frac{1}{k}−\frac{1}{j}\).

a. Apply the limit to both sides of the inequality in part c. to conclude that

\(\displaystyle T_k−γ≤\frac{1}{k}.\)

e. Estimate \(\displaystyle γ\) to an accuracy of within **0.001**.

## Key Concepts

- Given the infinite series

\(\displaystyle \sum_{n=1}^∞a_n=a_1+a_2+a_3+⋯\)

and the corresponding sequence of partial sums \(\displaystyle {S_k}\) where

\(\displaystyle S_k=\sum_{n=1}^ka_n=a_1+a_2+a_3+⋯+a_k\),

the series converges if and only if the sequence \(\displaystyle {S_k}\) converges.

- The geometric series \(\displaystyle \sum^∞_{n=1}ar^{n−1}\) converges if \(\displaystyle |r|<1\) and diverges if \(\displaystyle |r|≥1.\) For \(\displaystyle |r|<1,\)

\(\displaystyle \sum_{n=1}^∞ar^{n−1}=\frac{a}{1−r}\).

- The harmonic series

\(\displaystyle \sum_{n=1}^∞\frac{1}{n}=1+\frac{1}{2}+\frac{1}{3}+⋯\)

diverges.

- A series of the form \(\displaystyle \sum_{n=1}^∞[b_n−b_{n+1}]=[b_1−b_2]+[b_2−b_3]+[b_3−b_4]+⋯+[b_n−b_{n+1}]+⋯\) is a telescoping series. The \(\displaystyle kth\) partial sum of this series is given by \(\displaystyle S_k=b_1−b_{k+1}\). The series will converge if and only if \(\displaystyle \lim_{k→∞}b_{k+1}|) exists. In that case,

\(\displaystyle \sum_{n=1}^∞[b_n−b_{n+1}]=b_1−\lim_{k→∞}(b_{k+1})\).

## Key Equations

**Harmonic series**

\(\displaystyle \sum_{n=1}^∞\frac{1}{n}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+⋯\)

**Sum of a geometric series**

\(\displaystyle \sum_{n=1}^∞ar^{n−1}=\frac{a}{1−r}\) for \(\displaystyle |r|<1\)

## Glossary

**convergence of a series**- a series converges if the sequence of partial sums for that series converges

**divergence of a series**- a series diverges if the sequence of partial sums for that series diverges

**geometric series**- a geometric series is a series that can be written in the form
\(\displaystyle \sum_{n=1}^∞ar^{n−1}=a+ar+ar^2+ar^3+⋯\)

**harmonic series**- the harmonic series takes the form
\(\displaystyle \sum_{n=1}^∞\frac{1}{n}=1+\frac{1}{2}+\frac{1}{3}+⋯\)

**infinite series**- an infinite series is an expression of the form
\(\displaystyle a_1+a_2+a_3+⋯=\sum_{n=1}^∞a_n\)

**partial sum**-
the \(\displaystyle kth\) partial sum of the infinite series \(\displaystyle \sum^∞_{n=1}a_n\) is the finite sum

\(\displaystyle S_k=\sum_{n=1}^ka_n=a_1+a_2+a_3+⋯+a_k\)

**telescoping series**- a telescoping series is one in which most of the terms cancel in each of the partial sums

### Contributors

Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org.