Sequences

Abigail Huettig

6 Sequences

Abbott provides us with the following definition for a sequence [1].

Definition I.12
A sequence is a function whose domain is $\mathbb{N}$ .

So, a sequence $\{a_n\}$ defined by a function $f(n)$ is really just a number pattern where $f(n)$ is the $n$ th term $a_n$ in the pattern. In calculus, we are interested in whether the sequence converges or diverges (does not converge).

One way to find whether a sequence $\{a_n\}$ defined by $f(n)$ converges is to find

$\lim_{n\rightarrow\infty}f(n).$

If this limit exists, then the sequence not only converges but converges to that limit [15].

We also have a theorem that can be used to determine if a sequence converges when it is hard to find the limit or to know if it even exists. The theorem is stated by Larson and Edwards as follows [15].

Theorem I.13

If a sequence ${a_n}$ is bounded and monotonic, then it converges.

This theorem raises an important question. What do bounded and monotonic mean?

A sequence is bounded above when $a_n$ is less than or equal to a real number $M$ for all $n$ and bounded below when $a_n$ is greater than or equal to $M$ for all $n$ . A sequence is bounded when it is bounded above and below [15].

Monotonic means that either

$a_n\leq a_{n+1}\,\,\text{or}\,\, a_n\geq a_{n+1}$

for all $n$ [15]. That is, the sequence is either nondecreasing or nonincreasing.

We will not go through a formal proof here, but we will discuss why this theorem makes sense. Consider a nondecreasing, bounded sequence $\{a_n\}$ . That is,

$a_1\leq a_2\leq ...\leq M$

where $M$ is a real number. Every sequence must either converge or diverge, and a nondecreasing sequence diverges if and only if for every real number $x$ we can find $n$ such that $a_n\geq x$ . Because there exists a real number $x$ such that $a_n\leq M< x$ for all $n$ , $\{a_n\}$ does not diverge and must converge. Thus, if a sequence is nondecreasing and bounded, it must converge. Similar reasoning can be used for a nonincreasing, bounded sequence.

Let’s now look at a problem that demonstrates how to prove a sequence is monotone and bounded so that the theorem can be applied. This is a problem I completed for Calculus II and comes from Larson and Edwards [15].

Example 14
Does $a_n=\displaystyle\frac{5}{n}+8$ converge?

Solution
The terms of $\{a_n\}$ are

$13,\frac{21}{2},\frac{29}{3},....$

The terms appear to be decreasing but will the sequence continue this way? Because $n$ is positive, we have that for all $n$

$\begin{align*} \frac{1}{n}&>\frac{1}{n+1} \\ \frac{5}{n}+8 &> \frac{5}{n+1}+8 &&\text{Multiply by 5 and add 8}\\ a_n&>a_{n+1} \end{align*}$

Thus, $a_n$ is decreasing and consequently monotonic. Moreover, $a_n$ is bounded above by 13.

Now, we need to determine if $a_n$ is bounded below. Because $n$ is a natural number, we have

$\begin{align*} \frac{1}{n}&\geq0 \\ \frac{5}{n}+8 &\geq 8 &&\text{Multiply by 5 and add 8}\\ \end{align*}$

Thus, $a_n$ is bounded below by 8. By Theorem I.13, $a_n$ converges.

We are now going to begin our discussion of series, which builds upon the ideas of sequences.

6 Sequences

License

Share This Book