Sequences

In this section, we introduce sequences and define what it means for a sequence to converge or diverge. We show how to find limits of sequences that converge, often by using the properties of limits for functions discussed earlier. We close this section with the Monotone Convergence Theorem, a tool we can use to prove that certain types of sequences converge.

Terminology of Sequences

To work with this new topic, we need some new terms and definitions. First, an infinite sequence is an ordered list of numbers of the form

Each of the numbers in the sequence is called a term. The symbol $n$

is called the index variable for the sequence. We use the notation

to denote this sequence. A similar notation is used for sets, but a sequence is an ordered list, whereas a set is not ordered. Because a particular number $a_n$

exists for each positive integer $n,$

we can also define a sequence as a function whose domain is the set of positive integers.

Let’s consider the infinite, ordered list

This is a sequence in which the first, second, and third terms are given by $a_1=2,$

and $a_3=8.$

You can probably see that the terms in this sequence have the following pattern:

Assuming this pattern continues, we can write the $n\text{th}$

term in the sequence by the explicit formula $a_n=2^n.$

Using this notation, we can write this sequence as

Alternatively, we can describe this sequence in a different way. Since each term is twice the previous term, this sequence can be defined recursively by expressing the $n\text{th}$

term $a_n$

in terms of the previous term $a_{n-1}.$

In particular, we can define this sequence as the sequence $\left\{a_n\right\}$

where $a_1=2$

and for all $n\ge 2,$

each term $a_n$

is defined by the recurrence relation$a_n=2a_{n-1}.$

Note that the index does not have to start at $n=1$

but could start with other integers. For example, a sequence given by the explicit formula $a_n=f(n)$

could start at $n=0,$

in which case the sequence would be

Similarly, for a sequence defined by a recurrence relation, the term $a_0$

may be given explicitly, and the terms $a_n$

for $n\ge 1$

may be defined in terms of $a_{n-1}.$

Since a sequence $\left\{a_n\right\}$

has exactly one value for each positive integer $n,$

it can be described as a function whose domain is the set of positive integers. As a result, it makes sense to discuss the graph of a sequence. The graph of a sequence $\left\{a_n\right\}$

consists of all points $(n,a_n)$

for all positive integers $n.$

[link] shows the graph of $\left\{2^n\right\}.$

Two types of sequences occur often and are given special names: arithmetic sequences and geometric sequences. In an arithmetic sequence, the difference between every pair of consecutive terms is the same. For example, consider the sequence

You can see that the difference between every consecutive pair of terms is $4.$

Assuming that this pattern continues, this sequence is an arithmetic sequence. It can be described by using the recurrence relation

Note that

Thus the sequence can also be described using the explicit formula

In general, an arithmetic sequence is any sequence of the form $a_n=cn+b.$

In a geometric sequence, the ratio of every pair of consecutive terms is the same. For example, consider the sequence

We see that the ratio of any term to the preceding term is $-\frac{1}{3}.$

Assuming this pattern continues, this sequence is a geometric sequence. It can be defined recursively as

Alternatively, since

we see that the sequence can be described by using the explicit formula

The sequence $\left\{2^n\right\}$

that we discussed earlier is a geometric sequence, where the ratio of any term to the previous term is $2.$

In general, a geometric sequence is any sequence of the form $a_n=c{r}^n.$

Limit of a Sequence

A fundamental question that arises regarding infinite sequences is the behavior of the terms as $n$

gets larger. Since a sequence is a function defined on the positive integers, it makes sense to discuss the limit of the terms as $n\to \infty .$

For example, consider the following four sequences and their different behaviors as $n\to \infty$

(see [link]):

1. $\left\{1+3n\right\}=\{4,7,10,13\text{,…}\}.$

   The terms

   $1+3n$

   become arbitrarily large as

   $n\to \infty .$

   In this case, we say that

   $1+3n\to \infty$

   as

   $n\to \infty .$
2. $\left\{1-{\left(\frac{1}{2}\right)}^n\right\}=\left\{\frac{1}{2},\frac{3}{4},\frac{7}{8},\frac{15}{16}\text{,…}\right\}.$

   The terms

   $1-{\left(\frac{1}{2}\right)}^n\to 1$

   as

   $n\to \infty .$
3. $\left\{{\left(\mathrm{-1}\right)}^n\right\}=\{\text{−}1,1,\mathrm{-1},1\text{,…}\}.$

   The terms alternate but do not approach one single value as

   $n\to \infty .$
4. $\left\{\frac{{\left(\mathrm{-1}\right)}^n}{n}\right\}=\left\{\mathrm{-1},\frac{1}{2},-\frac{1}{3},\frac{1}{4}\text{,…}\right\}.$

   The terms alternate for this sequence as well, but

   $\frac{{\left(\mathrm{-1}\right)}^n}{n}\to 0$

   as

   $n\to \infty .$

From these examples, we see several possibilities for the behavior of the terms of a sequence as $n\to \infty .$

In two of the sequences, the terms approach a finite number as $n\to \infty .$

In the other two sequences, the terms do not. If the terms of a sequence approach a finite number $L$

as $n\to \infty ,$

we say that the sequence is a convergent sequence and the real number $L$

is the limit of the sequence. We can give an informal definition here.

From [link], we see that the terms in the sequence $\left\{1-{\left(\frac{1}{2}\right)}^n\right\}$

are becoming arbitrarily close to $1$

as $n$

becomes very large. We conclude that $\left\{1-{\left(\frac{1}{2}\right)}^n\right\}$

is a convergent sequence and its limit is $1.$

In contrast, from [link], we see that the terms in the sequence $1+3n$

are not approaching a finite number as $n$

becomes larger. We say that $\{1+3n\}$

is a divergent sequence.

In the informal definition for the limit of a sequence, we used the terms “arbitrarily close” and “sufficiently large.” Although these phrases help illustrate the meaning of a converging sequence, they are somewhat vague. To be more precise, we now present the more formal definition of limit for a sequence and show these ideas graphically in [link].

We remark that the convergence or divergence of a sequence $\left\{a_n\right\}$

depends only on what happens to the terms $a_n$

as $n\to \infty .$

Therefore, if a finite number of terms $b_1,b_2\text{,…},b_N$

are placed before $a_1$

to create a new sequence

this new sequence will converge if $\left\{a_n\right\}$

converges and diverge if $\left\{a_n\right\}$

diverges. Further, if the sequence $\left\{a_n\right\}$

converges to $L,$

this new sequence will also converge to $L.$

As defined above, if a sequence does not converge, it is said to be a divergent sequence. For example, the sequences $\{1+3n\}$

and $\left\{{\left(\mathrm{-1}\right)}^n\right\}$

shown in [link] diverge. However, different sequences can diverge in different ways. The sequence $\left\{{\left(\mathrm{-1}\right)}^n\right\}$

diverges because the terms alternate between $1$

and $\mathrm{-1},$

but do not approach one value as $n\to \infty .$

On the other hand, the sequence $\{1+3n\}$

diverges because the terms $1+3n\to \infty$

as $n\to \infty .$

We say the sequence $\{1+3n\}$

diverges to infinity and write $\underset{n\to \infty }{\text{lim}}(1+3n)=\infty .$

It is important to recognize that this notation does not imply the limit of the sequence $\{1+3n\}$

exists. The sequence is, in fact, divergent. Writing that the limit is infinity is intended only to provide more information about why the sequence is divergent. A sequence can also diverge to negative infinity. For example, the sequence $\{\text{−}5n+2\}$

diverges to negative infinity because $\mathrm{-5}n+2\to \text{−}\infty$

as $n\to \text{−}\infty .$

We write this as $\underset{n\to \infty }{\text{lim}}\left(\mathrm{-5}n+2\right)=\to \text{−}\infty .$

Because a sequence is a function whose domain is the set of positive integers, we can use properties of limits of functions to determine whether a sequence converges. For example, consider a sequence $\left\{a_n\right\}$

and a related function $f$

defined on all positive real numbers such that $f(n)=a_n$

for all integers $n\ge 1.$

Since the domain of the sequence is a subset of the domain of $f,$

if $\underset{x\to \infty }{\text{lim}}f(x)$

exists, then the sequence converges and has the same limit. For example, consider the sequence $\left\{\frac{1}{n}\right\}$

and the related function $f\left(x\right)=\frac{1}{x}.$

Since the function $f$

defined on all real numbers $x>0$

satisfies $f\left(x\right)=\frac{1}{x}\to 0$

as $x\to \infty ,$

the sequence $\left\{\frac{1}{n}\right\}$

must satisfy $\frac{1}{n}\to 0$

as $n\to \infty .$

We can use this theorem to evaluate $\underset{n\to \infty }{\text{lim}}{r}^n$

for $0\le r\le 1.$

For example, consider the sequence $\left\{{\left(1\text{/}2\right)}^n\right\}$

and the related exponential function $f(x)={\left(1\text{/}2\right)}^x.$

Since $\underset{x\to \infty }{\text{lim}}{\left(1\text{/}2\right)}^x=0,$

we conclude that the sequence $\left\{{\left(1\text{/}2\right)}^n\right\}$

converges and its limit is $0.$

Similarly, for any real number $r$

such that $0\le r<1,$

and therefore the sequence $\left\{r^n\right\}$

converges. On the other hand, if $r=1,$

then $\underset{x\to \infty }{\text{lim}}{r}^x=1,$

and therefore the limit of the sequence $\left\{1^n\right\}$

is $1.$

If $r>1,$

and therefore we cannot apply this theorem. However, in this case, just as the function $r^x$

grows without bound as $n\to \infty ,$

the terms $r^n$

in the sequence become arbitrarily large as $n\to \infty ,$

and we conclude that the sequence $\left\{r^n\right\}$

diverges to infinity if $r>1.$

We summarize these results regarding the geometric sequence $\left\{r^n\right\}\text{:}$

Later in this section we consider the case when $r<0.$

We now consider slightly more complicated sequences. For example, consider the sequence $\{{(2\text{/}3)}^n+{(1\text{/}4)}^n\}.$

The terms in this sequence are more complicated than other sequences we have discussed, but luckily the limit of this sequence is determined by the limits of the two sequences $\left\{{(2\text{/}3)}^n\right\}$

and $\left\{{(1\text{/}4)}^n\right\}.$

As we describe in the following algebraic limit laws, since $\left\{{(2\text{/}3)}^n\right\}$

and ${\{1\text{/}4)}^n\}$

both converge to $0,$

the sequence $\{{(2\text{/}3)}^n+{(1\text{/}4)}^n\}$

converges to $0+0=0.$

Just as we were able to evaluate a limit involving an algebraic combination of functions $f$

and $g$

by looking at the limits of $f$

and $g$

(see Introduction to Limits), we are able to evaluate the limit of a sequence whose terms are algebraic combinations of $a_n$

and $b_n$

by evaluating the limits of $\left\{a_n\right\}$

and $\left\{b_n\right\}.$

Proof

We prove part iii.

Let $ϵ>0.$

Since $\underset{n\to \infty }{\text{lim}}{a}_n=A,$

there exists a constant positive integer $N_1$

such that for all $n\ge N_1.$

Since $\underset{n\to \infty }{\text{lim}}{b}_n=B,$

there exists a constant $N_2$

such that $\backslash |b_n-B\backslash |<\epsilon \text{/}2$

for all $n\ge N_2.$

Let $N$

be the largest of $N_1$

and $N_2.$

Therefore, for all $n\ge N,$

□

The algebraic limit laws allow us to evaluate limits for many sequences. For example, consider the sequence $\left\{\frac{1}{n^2}\right\}.$

As shown earlier, $\underset{n\to \infty }{\text{lim}}1\text{/}n=0.$

Similarly, for any positive integer $k,$

we can conclude that

In the next example, we make use of this fact along with the limit laws to evaluate limits for other sequences.

Recall that if $f$

is a continuous function at a value $L,$

then $f\left(x\right)\to f(L)$

as $x\to L.$

This idea applies to sequences as well. Suppose a sequence $a_n\to L,$

and a function $f$

is continuous at $L.$

Then $f\left(a_n\right)\to f\left(L\right).$

This property often enables us to find limits for complicated sequences. For example, consider the sequence $\sqrt{5-\frac{3}{n^2}}.$

From [link]a. we know the sequence $5-\frac{3}{n^2}\to 5.$

Since $\sqrt{x}$

is a continuous function at $x=5,$

Proof

Let $ϵ>0.$

Since $f$

is continuous at $L,$

there exists $\delta >0$

such that $\backslash |f\left(x\right)-f\left(L\right)\backslash |<\epsilon$

if $\backslash |x-L\backslash |<\delta .$

Since the sequence $\left\{a_n\right\}$

converges to $L,$

there exists $N$

such that $\backslash |a_n-L\backslash |<\delta$

for all $n\ge N.$

Therefore, for all $n\ge N,$

which implies $\backslash |f(a_n)\text{−}f(L)\backslash |<\epsilon .$

We conclude that the sequence $\left\{f\left(a_n\right)\right\}$

converges to $f\left(L\right).$

□

Another theorem involving limits of sequences is an extension of the Squeeze Theorem for limits discussed in Introduction to Limits.

Proof

Let $\epsilon >0.$

Since the sequence $\left\{a_n\right\}$

converges to $L,$

there exists an integer $N_1$

such that $\backslash |a_n-L\backslash |<\epsilon$

for all $n\ge N_1.$

Similarly, since $\left\{c_n\right\}$

converges to $L,$

there exists an integer $N_2$

such that $\backslash |c_n-L\backslash |<\epsilon$

for all $n\ge N_2.$

By assumption, there exists an integer $N$

such that $a_n\le b_n\le c_n$

for all $n\ge N.$

Let $M$

be the largest of $N_1,N_2,$

and $N.$

We must show that $\backslash |b_n-L\backslash |<\epsilon$

for all $n\ge M.$

For all $n\ge M,$

Therefore, $\text{−}\epsilon <b_n-L<\epsilon ,$

and we conclude that $\backslash |b_n-L\backslash |<\epsilon$

for all $n\ge M,$

and we conclude that the sequence $\left\{b_n\right\}$

converges to $L.$

□

Using the idea from [link]b. we conclude that $r^n\to 0$

for any real number $r$

such that $\mathrm{-1}<r<0.$

If $r<\text{−}1,$

the sequence $\left\{r^n\right\}$

diverges because the terms oscillate and become arbitrarily large in magnitude. If $r=\mathrm{-1},$

the sequence $\left\{r^n\right\}=\left\{{\left(\mathrm{-1}\right)}^n\right\}$

diverges, as discussed earlier. Here is a summary of the properties for geometric sequences.

Bounded Sequences

We now turn our attention to one of the most important theorems involving sequences: the Monotone Convergence Theorem. Before stating the theorem, we need to introduce some terminology and motivation. We begin by defining what it means for a sequence to be bounded.

For example, the sequence $\left\{1\text{/}n\right\}$

is bounded above because $1\text{/}n\le 1$

for all positive integers $n.$

It is also bounded below because $1\text{/}n\ge 0$

for all positive integers n. Therefore, $\left\{1\text{/}n\right\}$

is a bounded sequence. On the other hand, consider the sequence $\left\{2^n\right\}.$

Because $2^n\ge 2$

for all $n\ge 1,$

the sequence is bounded below. However, the sequence is not bounded above. Therefore, $\left\{2^n\right\}$

is an unbounded sequence.

We now discuss the relationship between boundedness and convergence. Suppose a sequence $\left\{a_n\right\}$

is unbounded. Then it is not bounded above, or not bounded below, or both. In either case, there are terms $a_n$

that are arbitrarily large in magnitude as $n$

gets larger. As a result, the sequence $\left\{a_n\right\}$

cannot converge. Therefore, being bounded is a necessary condition for a sequence to converge.

Note that a sequence being bounded is not a sufficient condition for a sequence to converge. For example, the sequence $\left\{{\left(\mathrm{-1}\right)}^n\right\}$

is bounded, but the sequence diverges because the sequence oscillates between $1$

and $\mathrm{-1}$

and never approaches a finite number. We now discuss a sufficient (but not necessary) condition for a bounded sequence to converge.

Consider a bounded sequence $\left\{a_n\right\}.$

Suppose the sequence $\left\{a_n\right\}$

is increasing. That is, $a_1\le a_2\le a_3\text{…}.$

Since the sequence is increasing, the terms are not oscillating. Therefore, there are two possibilities. The sequence could diverge to infinity, or it could converge. However, since the sequence is bounded, it is bounded above and the sequence cannot diverge to infinity. We conclude that $\left\{a_n\right\}$

converges. For example, consider the sequence

Since this sequence is increasing and bounded above, it converges. Next, consider the sequence

Even though the sequence is not increasing for all values of $n,$

we see that $\mathrm{-1}\text{/}2<\text{−}1\text{/}3<\text{−}1\text{/}4<\text{⋯}.$

Therefore, starting with the eighth term, $a_8=\mathrm{-1}\text{/}2,$

the sequence is increasing. In this case, we say the sequence is eventually increasing. Since the sequence is bounded above, it converges. It is also true that if a sequence is decreasing (or eventually decreasing) and bounded below, it also converges.

We now have the necessary definitions to state the Monotone Convergence Theorem, which gives a sufficient condition for convergence of a sequence.

The proof of this theorem is beyond the scope of this text. Instead, we provide a graph to show intuitively why this theorem makes sense ([link]).

In the following example, we show how the Monotone Convergence Theorem can be used to prove convergence of a sequence.

Key Concepts

• To determine the convergence of a sequence given by an explicit formula $a_n=f(n),$

  we use the properties of limits for functions.

• If $\left\{a_n\right\}$

  and

  $\left\{b_n\right\}$

  are convergent sequences that converge to

  $A$

  and

  $B,$

  respectively, and

  $c$

  is any real number, then the sequence

  $\left\{c{a}_n\right\}$

  converges to

  $c·A,$

  the sequences

  $\{a_n\pm b_n\}$

  converge to

  $A\pm B,$

  the sequence

  $\{a_n·b_n\}$

  converges to

  $A·B,$

  and the sequence

  $\left\{a_n\text{/}{b}_n\right\}$

  converges to

  $A\text{/}B,$

  provided

  $B\ne 0.$
• If a sequence is bounded and monotone, then it converges, but not all convergent sequences are monotone.
• If a sequence is unbounded, it diverges, but not all divergent sequences are unbounded.
• The geometric sequence $\left\{r^n\right\}$

  converges if and only if

  $\backslash |r\backslash |<1$

  or

  $r=1.$