Sequences

Abigail Huettig

20 Sequences

Recall that a sequence is defined as a function with the natural numbers as the domain [1]. When we discussed sequences in the Calculus section, we mentioned that our goal is to determine whether a sequence converges or diverges. If a sequence does converge, we want to know what numerical value it converges to. We also discussed what it means for a sequence to be bounded or monotonic, and we introduced the Monotone Convergence Theorem which says that if a sequence is both bounded and monotonic then it converges. Refer back to Part I: Chapter 6 for a more detailed discussion of these topics. In this section, we will begin by taking a look at Cauchy sequences.

Cauchy Sequences

When we discussed sequences in Calculus, we mentioned that they converged if the limit as $n$ approached infinity of $f(n)$ existed, where $f(n)$ is the function that defines the sequence. We will now introduce the formal definition of convergence as given by Abbott [1].

Definition IV.14
A sequence $(a_n)$ converges to a real number $a$ if, for every $\epsilon>0$ , there exists an $N\in \mathbb{N}$ such that whenever $n\geq N$ it follows that $|a_n-a|<\epsilon$ .

Now that we have defined convergence, we are ready for the definition of a Cauchy sequence given by Abbott as follows [1].

Definition IV.15
A sequence $(a_n)$ is called a Cauchy sequence if, for every $\epsilon>0$ , there exists an $N\in\mathbb{N}$ such that whenever $m,n\geq N$ it follows that $|a_n-a_m|<\epsilon.$

This definition looks very similar to the definition of convergence. What the definition of convergence is saying is that if a sequence converges then there exists a point after which every term of the sequence is within a distance of $\epsilon$ of a real number $a$ [1]. A sequence is Cauchy if there exists a point such that for any two terms $a_m$ and $a_n$ after that point, $a_m$ and $a_n$ are within a distance of $\epsilon$ of each other [1].

Let’s use this information to complete an exercise. The following problem is one I completed as homework for Introduction to Analysis [6].

Example 43
Show that the sequence $a_n=\frac{1}{n}$ is a Cauchy sequence.

Proof.
Let $\epsilon>0$ . Choose $N\in \mathbb{N}$ such that $N>\frac{2}{\epsilon}$ . Let $m,n\geq N$ where $m,n\in \mathbb{N}$ . So, $m>\frac{2}{\epsilon}$ and $n>\frac{2}{\epsilon}$ , which gives us the following.

$\begin{align*} m&>\frac{2}{\epsilon}\\ m\epsilon&>2\\ \epsilon&>\frac{2}{m}\\ \frac{\epsilon}{2}&>\frac{1}{m}\\ &=\left|-\frac{1}{m}\right| \end{align*}$

and

$\begin{align*} n&>\frac{2}{\epsilon} \\ n\epsilon&>2\\ \epsilon&>\frac{2}{n}\\ \frac{\epsilon}{2}&>\frac{1}{n}\\ &=\left|\frac{1}{n}\right| \end{align*}$

It then follows that

$\begin{align*} |a_n-a_m|&=\left|\frac{1}{n}-\frac{1}{m}\right|\\ &=\left|\frac{1}{n}+\left(-\frac{1}{m}\right)\right|\\ &\leq \left|\frac{1}{n}\right|+\left|-\frac{1}{m}\right| &&\text{Triangle Inequality}\\ &<\frac{\epsilon}{2}+\frac{\epsilon}{2}\\ &=\epsilon \end{align*}$

Thus, for $a_n=\frac{1}{n}$ , there exists $N\in\mathbb{N}$ such that when $m,n\geq N$ , $|a_n-a_m|0.$ Therefore, the sequence $a_n=\frac{1}{n}$ is a Cauchy sequence.

There is an interesting result relating to Cauchy sequences that helps us with convergence, known as the Cauchy Criterion. It is stated by Abbot as follows [1].

Theorem IV.16 (Cauchy Criterion)

A sequence converges if and only if it is a Cauchy sequence.

Now, we not only have the Monotone Convergence Theorem but we also have the Cauchy Criterion to help us determine whether a sequence converges. In the next section, we are going to see yet another way to determine convergence by looking at subsequences.

Subsequences and Convergence

Another way we can tell if a sequence converges or diverges is to look at how its subsequences behave. So, let’s start with the formal definition of a subsequence that comes from Abbott [1].

Definition IV.17
Let $(a_n)$ be a sequence of real numbers, and let $n_1<n_2<n_3<n_4<n_5<...$ be an increasing sequence of natural numbers. Then the sequence

$(a_{n_1},a_{n_2},a_{n_3}, a_{n_4}, a_{n_5})$

is called a subsequence of $(a_n)$ and is denoted by $(a_{n_k})$ , where $k\in \mathbb{N}$ indexes the subsequence.

A subsequence $(a_{n_k})$ of $(a_n)$ is a sequence that must satisfy three criteria.

Every term in a subsequence must appear in the original sequence.
A term should not appear more often in a subsequence than it does in the original sequence.
The terms of a subsequence must appear in the same order as the original sequence.

Now that we understand what a subsequence is, we are ready to look at a couple theorems related to them that are stated by Abbott as follows.

Theorem IV.18

Subsequences of a convergent sequence converge to the same limit as the original sequence.

Stating this theorem another way, we have, “if a sequence converges to a limit $L$ , then all of its subsequences converge to the same limit $L$ ,” and the contrapositive of this statement is known as the Divergence Criterion. That is, the Divergence Criterion says, “if at least two subsequences converge to different limits, then the original sequence diverges.” So, using Theorem IV.18 in the form of the Divergence Criterion, we can show that a sequence diverges by finding two subsequences that converge to different limits.

Theorem IV.19 (Bolzano-Weierstrass Theorem)

Every bounded sequence contains a convergent subsequence.

Notice that the theorem states that the sequence only has to be bounded. It is possible for a sequence not to converge and still contain a convergent subsequence. For example, the sequence

$a_n=(-1)^n={-1,1,-1,1,...}$

has two converging subsequences,

$a_{n_k}=1={1,1,1,1,...}$

and

$a_{n_k}=-1={-1,-1,-1,-1,...}.$

However, because the first subsequence converges to 1 and the other converges to -1, the original sequence does not converge by the Divergence Criterion.

Let’s now look at an example that shows another way Theorem IV.18 can be used. This problem comes from Abbott and is one I completed as homework for Introduction to Analysis [1]. We have not mentioned the Algebraic Limit Theorem, and we will not be discussing it here. But note that the Algebraic Limit Theorem states that

$\lim(a_nb_n)=ab$

where $\lim(a_n)=a$ and $\lim(b_n)=b$ [1].

Example 44
Show that $\lim(b^{\frac{1}{n}})$ exists for all $b\geq 1$ and find the value of the limit.

Proof.
If $b\geq 1$ and $b^{\frac{1}{n}}=b^1,b^{\frac{1}{2}},b^{\frac{1}{3}},...$ ,

$b^1\geq b^{\frac{1}{2}}\geq b^{\frac{1}{3}}\geq...\geq 1.$

Thus, $(b^{\frac{1}{n}})$ is monotone, bounded above by $b$ , and bounded below by 1. By the Monotone Convergence Theorem, $(b^{\frac{1}{n}})$ converges to a real number $\ell$ where $1\leq\ell\leq b$ .

Now consider the subsequence $(b^{\frac{1}{2n}})$ . $(b^{\frac{1}{2n}})$ also converges to $\ell$ by Theorem IV.18. Because $b^{\frac{1}{n}}= b^{\frac{1}{2n}}\cdot b^{\frac{1}{2n}}$ , we have

$\begin{align*} \ell&=\lim(b^{\frac{1}{n}})\\ &=\lim(b^{\frac{1}{2n}}\cdot b^{\frac{1}{2n}})\\ &=\ell^2 \end{align*}$

by the Algebraic Limit Theorem. Since $\ell=\ell^2$ , $\ell$ must be -1, 0, or 1. Because $1\leq\ell\leq b$ , $\ell=1$ . Therefore, $\lim(b^{\frac{1}{n}})=1$ for all $b\geq 1$ .

We have now looked at sequences from both a Calculus perspective and an Analysis perspective and seen how the different views interact with one another. Next, we will look at limits and continuity from the point of view of analysis and see how it compares to that of calcul

20 Sequences

Cauchy Sequences

Subsequences and Convergence

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