Divergence to infinity – "Math for Non-Geeks"

Until to now, we investigated when and how a sequence diverges. In this chapter, we will investigate divergent series. Not every divergent series behaves the same: some can be seen to diverge to $+\mathrm{\infty }$ or $-\mathrm{\infty }$ (so we can assign a formal limit to them) and some don't.

Motivation

Some sequences do not converge, but they unambiguously tend to $+\mathrm{\infty }$ or $-\mathrm{\infty }$. For instance consider the sequences $a_n=n$, $b_n=2^n$ and $c_n=-n+2\cdot (-1{)}^n$:

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  The sequence $a_n=n$ tends towards $+\mathrm{\infty }$.
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  The sequence $b_n=2^n$ tends towards $+\mathrm{\infty }$ even faster.
• 
  The sequence $c_n=-n+2\cdot (-1{)}^n$ tends towards $-\mathrm{\infty }$.

We will give a mathematical definition which classifies these sequences as diverging towards $+\mathrm{\infty }$ or $-\mathrm{\infty }$. Some other sequences may not diverge this way. For instance, consider $d_n=(-1{)}^n$ or $e_n=(-1{)}^n\cdot n$ .

The sequence $d_n=(-1{)}^n$ is bounded and can therefore neither tend to $+\mathrm{\infty }$, not to $-\mathrm{\infty }$ .

The sequence $e_n=(-1{)}^n\cdot n$ is unbounded, but contains parts (subsequences) tending towards $+\mathrm{\infty }$ and parts tending towards $-\mathrm{\infty }$ :

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  The alternating sequence $d_n=(-1{)}^n$ is bounded, so it cannot tend towards infinity.
• 
  The sequence $e_n=(-1{)}^n\cdot n$ is unbounded, but neither tends to $+\mathrm{\infty }$ nor to $-\mathrm{\infty }$.

Definition

We have observed some sequences, which tend towards $+\mathrm{\infty }$ or $-\mathrm{\infty }$ . How can we give a mathematically precise classification for this observation?

Let us start with divergence to $+\mathrm{\infty }$: "A sequence diverges to $+\mathrm{\infty }$" means that it grows larger than any number $S$, no matter how large (since $+\mathrm{\infty }$ is greater than any $S$). Even further, almost all sequence elements must be greater than $S$ . Or equivalently, we need a sequence element number $N$, such that any element coming after it is bigger than $S$ . Indeed, this is already sufficient as condition for divergence to $+\mathrm{\infty }$:

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

A sequence $(a_n{)}_{n\in \mathbb{N}}$ diverges to $+\mathrm{\infty }$, if for any $S\in ℝ$ almost all sequence elements are greater or equal to $S$ . That means, for all $S$ , there is an index $N$, such that $a_n\ge S$ for all $n\ge N$. In quantifier notation:

$\mathrm{\forall }S\in ℝ\mathrm{\exists }N\in \mathbb{N}\mathrm{\forall }n\ge N:a_n\ge S$

We can translate this quantifier notation piece by piece:

$\begin{array}{c}\underset{\text{For any real }S}{\underbrace{\mathrm{\forall }S\in ℝ}}\underset{\text{ there is a minimal index }N,}{\underbrace{\mathrm{\exists }N\in \mathbb{N}}}\underset{\text{such that for all indices }n\ge N}{\underbrace{\mathrm{\forall }n\ge N:}}\\ \underset{\text{ the element }{a}_n\text{ is bigger or equal }S}{\underbrace{a_n\ge S}}\end{array}$

Divergence to $-\mathrm{\infty }$ is just defined analogously:

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

A sequence $(a_n{)}_{n\in \mathbb{N}}$ diverges to $-\mathrm{\infty }$, if for any $s\in ℝ$ almost all sequence elements are lower or equal to $s$ . That means, for all $s$ , there is an index $N$, such that $a_n\le S$ for all $n\ge N$. In quantifier notation:

$\mathrm{\forall }s\in ℝ\mathrm{\exists }N\in \mathbb{N}\mathrm{\forall }n\ge N:a_n\le s$

Notation

For $(a_n{)}_{n\in \mathbb{N}}$ diverging to $+\mathrm{\infty }$ , we write

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

Analogously, for $(a_n{)}_{n\in \mathbb{N}}$ diverging to $-\mathrm{\infty }$ , we write

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

Examples

Example

$\begin{array}{cc}\underset{n\to \mathrm{\infty }}{lim}n& =+\mathrm{\infty }\\ \underset{n\to \mathrm{\infty }}{lim}{n}^2& =+\mathrm{\infty }\\ \underset{n\to \mathrm{\infty }}{lim}-n& =-\mathrm{\infty }\end{array}$

Example (geometric sequence)

We have seen in the article "unbounded sequences diverge" that the geometric sequence ${\left(q^n\right)}_{n\in \mathbb{N}}$ diverges for $|q|>1$. The reason is that for any $q>1$ and any real number $S$ , the inequality $q^n>S$ holds for almost all $n\in \mathbb{N}$ . Therefore, $\underset{n\to \mathrm{\infty }}{lim}{q}^n=\mathrm{\infty }$ for $q>1$. However, if $q\le -1$, there is neither a divergence to $+\mathrm{\infty }$ nor to $-\mathrm{\infty }$: For even $n$ , the sequence element $q^n$ is positive and for $n$ , it is negative. Hence, the geometric sequence ${\left(q^n\right)}_{n\in \mathbb{N}}$ cannot diverge for $q\le -1$ . For $|q|<1$ and $q=1$, it is convergent.

Improper convergence

The notation $\underset{n\to \mathrm{\infty }}{lim}{a}_n=+\mathrm{\infty }$ suggests that $(a_n{)}_{n\in \mathbb{N}}$ kind of "converges" to infinity. This is no convergence in the usual sense, since the symbolic expression $+\mathrm{\infty }$ is not a number. However, the divergence to $\pm \mathrm{\infty }$ has a lot in common with convergence to a number (except for boundedness):

Convergence	Divergence to $\pm \mathrm{\infty }$
In each $ϵ$-neighbourhood (interval), we can find almost all sequence elements.	In each interval $[S,\mathrm{\infty })$, we can find almost all sequence elements.
All subsequences converge to the same limit.	All subsequences also diverge to $\pm \mathrm{\infty }$.
Every convergent sequence is bounded.	Every sequence diverging to $\pm \mathrm{\infty }$ is unbounded.

Especially, some of the limit theorems hold true for $\pm \mathrm{\infty }$ and in some cases, one may treat a sequence diverging to $\pm \mathrm{\infty }$, similar to a convergent sequence. Sometimes, the divergence to $\pm \mathrm{\infty }$ is even called "improper convergence". However, always keep in mind that an improper convergence is still a divergence.

Warning

Some sequences are called "improperly convergent". Despite their name, they are actually divergent sequences. Some limit theorems still hold for improperly divergent sequences, for instance the product rule $\underset{n\to \mathrm{\infty }}{lim}(a_n\cdot b_n)=\underset{n\to \mathrm{\infty }}{lim}{a}_n\cdot \underset{n\to \mathrm{\infty }}{lim}{b}_n$ in case both sequences diverge. If one sequence converges and a second one diverges, some rules may also turn out wrong. For instance,

$1=\underset{n\to \mathrm{\infty }}{lim}(n\cdot \frac{1}{n})=\underset{n\to \mathrm{\infty }}{lim}n\cdot \underset{n\to \mathrm{\infty }}{lim}\frac{1}{n}=\mathrm{\infty }\cdot 0=0$

Applying the product rule to a product of a convergent and divergent sequence leads to the wrong result $1=0$. Always be careful, when using limit theorems for improperly convergent sequences!