4.1 Sequences & Convergence

A sequence is an infinitely long ordered list of real numbers. The precise definition is as follows.

Definition 4.1.

A sequence of real numbers is a function

f:\mathbb{N}\to\mathbb{R}.

We denote $f(n)$ as $x_{n}$ and write the sequence as

(x_{n})_{n},\quad(x_{n})_{n\in\mathbb{N}},\quad(x_{n})_{n=1}^{\infty},\quad(x_% {1},x_{2},\dots).

The elements of a sequence can be given by a formula for the $n$ -th term, a recurrence relation, or a pattern given by the first few terms.

(2,4,6,8,\dots)=(2n)_{n}\ \mathrm{i.e.}\ x_{n}=2n,\quad\left(1,\frac{1}{2},% \frac{1}{3},\frac{1}{4},\dots\right)=\left(\frac{1}{n}\right)_{n}\ \mathrm{i.e% .}\ x_{n}=\frac{1}{n}.

(4.1)

4.1.1 Defining Convergence

Some sequences converge to a fixed value, or limit, meaning that as $n$ gets larger and larger, $x_{n}$ gets closer and closer to the limit. An important question to ask is how do we define if a sequence converges or not, or how do we find the limit if it is only reached after an infinite amount of steps? For example, the sequence $\left(\frac{1}{n}\right)_{n}$ from above intuitively converges to 0 as $n$ tends to infinity, but how can we prove this mathematically?

Definition 4.2.

Let $(x_{n})_{n}$ be a sequence of real numbers and let $x\in\mathbb{R}$ .

1.

$(x_{n})_{n}$ is a convergent sequence with limit $x$ if

$\forall\varepsilon>0,\ \exists N\in\mathbb{N}\ :\ n\geq N\implies\lvert x_{n}-% x\rvert\leq\varepsilon.$ (4.2)

In english this says that ‘for all $\varepsilon>0$ , there exists a natural number $N$ such that for $n\geq N$ the distance between $x_{n}$ and the limit is less than $\varepsilon$ ’. We say $(x_{n})_{n}$ tends to $x$ as $n$ tends to infinity.
2.

$(x_{n})_{n}$ tends to infinity as $n$ tends to infinity if

$\forall K>0,\ \exists N\in\mathbb{N}\ :\ n\geq N\implies x_{n}\geq K.$ (4.3)
3.

Likewise, $(x_{n})_{n}$ tends to minus infinity as $n$ tends to infinity if

$\forall K>0,\ \exists N\in\mathbb{N}\ :\ n\geq N\implies x_{n}\leq-K.$ (4.4)
4.

A sequence is divergent if the limit does not exist or it tends to plus/minus infinity.

Let’s take a closer look at definition 4.2.1. Being unfamiliar with techniques in analysis, it might look very confusing, but it is actually quite a simple and clever idea. The key thing to realise is that converging sequences never actually reach their limit (unless they are constant), so in order to capture this idea of reaching a limit after infinitely many terms in concrete mathematics, we introduce this idea of bounding the elements of the sequence within a certain distance of the limit. Therefore, what this definition is really saying is that no matter how small epsilon is, there exists a step in the sequence such that from that point onwards all values in the sequence are inside the interval $(x-\varepsilon,x+\varepsilon)$ . Once this idea is understood, everything in analysis starts to make sense! Let us now look at a few examples.

Example 4.1.

Claim: The sequence $\left(\frac{1}{n}\right)_{n}$ tends to 0 as $n$ tends to infinity.
Proof. Let $\varepsilon>0$ and choose $N>\frac{1}{\varepsilon}$ . Then for $n\geq N$ we have

\lvert x_{n}-x\rvert=\left\lvert\frac{1}{n}-0\right\rvert=\frac{1}{n}\leq\frac% {1}{N}<\varepsilon.

(4.5)

Example 4.2.

Claim: $\left(\frac{1}{n^{2}}\right)_{n}\to 0$ as $n\to\infty$ .
Proof. Let $\varepsilon>0$ and choose $N>\frac{1}{\sqrt{\varepsilon}}$ . Then for $n\geq N$ we have

\lvert x_{n}-x\rvert=\left\lvert\frac{1}{n^{2}}-0\right\rvert=\frac{1}{n^{2}}% \leq\frac{1}{N^{2}}<\varepsilon.

(4.6)

It is important to note that we do not have to always use the most optimal value for $N$ . In this proof, choosing $N>\frac{1}{\varepsilon}$ would have been sufficient because $\frac{1}{N^{2}}\leq\frac{1}{N}$ .

Example 4.3.

Claim: $x_{n}=\frac{2n+1}{3n+5}$ , $(x_{n})_{n}\to\frac{3}{2}$ as $n\to\infty$ .
Proof. Let $\varepsilon>0$ and choose $N>\frac{7}{9}\varepsilon$ . Then for $n\geq N$ we have

$\displaystyle\lvert x_{n}-x\rvert$	$\displaystyle=\left\lvert\frac{2n+1}{3n+5}-\frac{2}{3}\right\rvert=\left\lvert% \frac{3(2n+1)-2(3n+5)}{3(3n+5)}\right\rvert$	(4.7)
	$\displaystyle=\left\lvert\frac{6n-6n+3-10}{9n+15}\right\rvert=\left\lvert\frac% {-7}{9n+15}\right\rvert=\frac{7}{9n+15}$	(4.8)
	$\displaystyle\leq\frac{7}{9N+15}\leq\frac{7}{9N}<\varepsilon.$	(4.9)

Example 4.4.

Claim: $(n^{3})_{n}\to\infty$ as $n\to\infty$ .
Proof. Let $K>0$ and choose $N>\sqrt[3]{K}$ . Then for $n\geq N$ we have

x_{n}=n^{3}\geq N^{3}>K.

(4.10)

Example 4.5.

Claim: $x_{n}=(-1)^{n}$ , $(x_{n})_{n}$ is divergent.
Proof. Suppose by way of contradiction that there is a limit $x\in\mathbb{R}$ .
Let $\varepsilon=\frac{1}{2}$ , then since the limit exists there exists $N$ such that $n\geq N$ implies $\lvert x_{n}-x\rvert<\frac{1}{2}$ .
Since there are arbitrarily large even numbers, there is always some $n\geq N$ (for example $n=2N$ ) such that $x_{n}=1$ , so $\lvert 1-x\rvert<\frac{1}{2}\implies x>\frac{1}{2}$ .
Also, since there are arbitrarily large odd numbers, there is always some $n\geq N$ ( $n=2N+1$ ) such that $x_{n}=-1$ , so $\lvert-1-x\rvert<\frac{1}{2}\implies x<-\frac{1}{2}$ .
But then $\frac{1}{2}<x<-\frac{1}{2}$ , which is impossible, so the limit cannot exist.

4.1.2 Standard Sequences

Theorem 4.1 (Standard Sequences)

Let $a\in\mathbb{R}$ .

(i)

Let $x_{n}=a$ , then $(x_{n})_{n}\to a$ as $n\to\infty$ (constant sequence).
(ii)

If $\lvert a\rvert<1$ , then $(a^{n})_{n}\to 0$ as $n\to\infty$ .
(iii)

If $a>1$ , then $(a^{n})_{n}\to\infty$ as $n\to\infty$ .

Proof.

(i)

Let $\varepsilon>0$ . Then given $N\in\mathbb{N}$ , $n\geq N$ implies

$\lvert x_{n}-x\rvert=\lvert x-x\rvert=0<\varepsilon.$ (4.11)
(ii)

Let $\varepsilon>0$ and choose $N>\frac{\ln{\varepsilon}}{\ln{\lvert a\rvert}}$ . Then for $n\geq N$ we have

$\lvert x_{n}-x\rvert=\lvert a^{n}-0\rvert=\lvert a^{n}\rvert\leq\lvert a^{N}% \rvert<\varepsilon.$ (4.12)
(iii)

Let $K>0$ and choose $N>\frac{\ln{K}}{\ln{a}}$ . Then for $n\geq N$ we have

$a^{n}\geq a^{N}>K.$ (4.13)

∎