Convergence and divergence of sequences

Consider the sequence $a_1,a_2,a_3,\cdots$ where each $a_i$ is a real number, and there is a real number $a_n$ for each natural number $n$. We usually denote such a sequence by $\{a_n\}$ or $\{a_n{\}}_{n=1}^{\mathrm{\infty }}$. Note that a sequence is not just a set, its order matters. $a_n$ may ``converge'' to some real number as $n$ becomes larger and larger. But what does that mean?

Definition (Limits)

Let $\{a_n\}$ be a sequence. If $a_n$ approaches arbitrarily close to a constant value $\alpha$ as $n$ becomes arbitrarily large, we call this $\alpha$ the limit of the sequence $\{a_n\}$ and write

$$\underset{n\to \mathrm{\infty }}{lim}{a}_n=\alpha$$

or

$$a_n\to \alpha \text{ as }n\to \mathrm{\infty },$$

and we say the sequence $\{a_n\}$ converges to $\alpha$. In this case, we also say ``the limit of $\{a_n\}$ is $\alpha$.''

If the sequence $\{a_n\}$ does not converge, we say that the sequence diverges.

Remark. When we say ``$a_n$ approaches to $\alpha$ as $n\to \mathrm{\infty }$'', it means $|a_n-\alpha |$ becomes smaller as $n$ becomes larger. But we will make this notion more precise below. □

Example. Let $\{a_n\}$ be a sequence with each term defined by $a_n=\frac{(-1{)}^n}{n}$ for each $n\in \mathbb{N}$. Then the sequence is

$$-1,\frac{1}{2},-\frac{1}{3},\frac{1}{4},-\frac{1}{5},\cdots .$$

□

Example. Let $\{a_n\}$ be a sequence defined by $a_n=3n-1$. We have

$$2,5,8,11,14,\cdots .$$

The numbers become arbitrarily large and positive numbers. Thus, this sequence diverges. □

In cases such as this last example, we say the sequence $\{a_n\}$ diverges to the positive infinity and write

$$\underset{n\to \mathrm{\infty }}{lim}{a}_n=\mathrm{\infty }$$

or

$$a_n\to \mathrm{\infty }(n\to \mathrm{\infty }).$$

Example. Let $\{a_n\}$ be defined by $a_n=-n^2$ so we have

$$-1,-4,-9,-16,-25,\cdots .$$

The numbers become arbitrarily large negative numbers. So this sequence diverges. □

As in this example, when a sequence diverges to arbitrarily large negative values, we say the sequence $\{a_n\}$ diverges to the negative infinity, and write

$$\underset{n\to \mathrm{\infty }}{lim}{a}_n=-\mathrm{\infty }$$

or

$$a_n\to -\mathrm{\infty }(n\to \mathrm{\infty }).$$

Example. Let $\{a_n\}$ be defined by $a_n=(-1{)}^n$. We have

$$-1,1,-1,1,\cdots$$

and this sequence diverges, but neither to the positive nor negative infinities. □

What do we exactly mean by convergence? Consider the sequence $a_n=\frac{(-1{)}^n}{n}$.

$$\begin{array}{ccccccccc}n& 1& 2& \cdots & 10& 100& 1,000& 10,000& \cdots \\ a_n& -1& 0.5& \cdots & 0.1& 0.01& 0.001& 0.0001& \cdots \end{array}$$

$\underset{n\to \mathrm{\infty }}{lim}{a}_n=0$ means that $|a_n|=|a_n-0|$ becomes arbitrarily small as $n$ becomes larger. For example,

• For $n>100$, $|a_n-0|<0.01$.
• For $n>1,000$, $|a_n-0|<0.001$.

Generalizing these observations, for any arbitrarily small positive number, say $\epsilon =0.0001$, we can always find some natural number, say $N=100,000$, such that if $n>N$, then $|a_n-0|<\epsilon$. Hence the following definition.

Definition (Convergence of a sequence)

The sequence $\{a_n\}$ is said to converge to a real number $\alpha$ if and only if the following condition holds:

• For any $\epsilon >0$, there exists $N\in \mathbb{N}$ such that for any $n\in \mathbb{N}$, if $n\ge N$, then $|a_n-\alpha |<\epsilon$.

(Here, we implicitly assume $\epsilon \in \mathbb{R}$.)

Remark. We call this type of argument using $\epsilon$ and $N$ the ``$\epsilon -N$ argument.'' □

Remark. In a logical form, the above condition for convergence can be expressed as

$$\mathrm{\forall }\epsilon >0,\mathrm{\exists }N\in \mathbb{N},\mathrm{\forall }n\in \mathbb{N}(n\ge N⟹|a_n-\alpha |<\epsilon ).$$

□

Example. If $a_n=\frac{n}{n+1}$, then $\{a_n\}$ converges to $\alpha =1$. For example, let $\epsilon =0.01$. If $|a_n-\alpha |<\epsilon$, then

$$1-\frac{n}{n+1}<0.01.$$

By solving this, we have

$$n>99.$$

Therefore, if we set $N=100$, then for any $n\ge N$, we have $|a_n-\alpha |<\epsilon$.

The same procedure can be applied for any values of $\epsilon >0$. □

Example. Let us prove that

$$\underset{n\to \mathrm{\infty }}{lim}\frac{1}{n}=0.$$

By Archimedes' principle, for any given $\epsilon >0$, we can find a natural number $N$ such that $\epsilon N>1$, and hence $\frac{1}{N}<\epsilon$. For $n\in \mathbb{N}$, if $n\ge N$, then $\frac{1}{n}\le \frac{1}{N}$, so that

$$|\frac{1}{n}-0|=\frac{1}{n}\le \frac{1}{N}<\epsilon .$$

□

Remark. Note that the choice of ``$N$'' depends on the value of $\epsilon$. For each $\epsilon >0$, we choose an appropriate $N\in \mathbb{N}$. We cannot choose one $N\in \mathbb{N}$ for all possible values of $\epsilon$. If that's the case, then we should have $\frac{1}{N}<\epsilon$ for any $\epsilon >0$, which implies $\frac{1}{N}=0$, which is nonsense.

Example. Let us prove that $a_n=(-1{)}^n$ does not converge by using the $\epsilon -N$ argument. 

We prove it by contradiction. 

Suppose that $\{a_n\}$ converges to some real number $\alpha$. Let's pick $\epsilon =1$. There should exist some $N\in \mathbb{N}$ such that, if $n\ge N$, then

$$|a_n-\alpha |<1.$$

When $n$ is even, $a_n=1$ so $|1-\alpha |<1$, in particular,

$$1-\alpha <1.$$

This implies that $\alpha >0$.

When $n$ is odd, $a_n=-1$ so $|-1-\alpha |<1$, in particular,

$$1+\alpha <1.$$

This implies $\alpha <0$.

Therefore $\alpha >0$ and $\alpha <0$, which is a contradiction. Hence $\{a_n\}$ does not converge. □

Definition (Divergence to $\pm \mathrm{\infty }$)

1. The sequence $\{a_n\}$ is said to diverge to the positive infinity, denoted $\underset{n\to \mathrm{\infty }}{lim}{a}_n=+\mathrm{\infty }$, if the following condition is satisfied.
• For any $M\in \mathbb{R}$, there exists $N\in \mathbb{N}$ such that for any $n\in \mathbb{N}$, if $n\ge N$, then $a_n>M$.
Or, in a logical form, $$\mathrm{\forall }M\in \mathbb{R},\mathrm{\exists }N\in \mathbb{N},\mathrm{\forall }n\in \mathbb{N}(n\ge N⟹a_n>M).$$
2. The sequence $\{a_n\}$ is said to diverge to the negative infinity, denoted $\underset{n\to \mathrm{\infty }}{lim}{a}_n=-\mathrm{\infty }$, if the following condition is satisfied.
• For any $M\in \mathbb{R}$, there exists $N\in \mathbb{N}$ such that for any $n\in \mathbb{N}$, if $n\ge N$, then $a_n<M$.
Or, in a logical form, $$\mathrm{\forall }M\in \mathbb{R},\mathrm{\exists }N\in \mathbb{N},\mathrm{\forall }n\in \mathbb{N}(n\ge N⟹a_n<M).$$

Example. Consider the sequence $\{3n^2\}$. For a given arbitrary positive real number $M>0$, let $N=\left[\sqrt{\frac{M}{3}}\right]+1$ where $\left[x\right]$ indicates the integer part of $x$. Note, in particular, that $N>\sqrt{\frac{M}{3}}$. Now, suppose $n\ge N$. We have $3n^2\ge 3N^2>M$. Therefore, the sequence $\{3n^2\}$ diverges to $+\mathrm{\infty }$. □