# Math for Non-Geeks/Divergence to infinity

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 $+\infty$ or $-\infty$ (so we can assign a formal limit to them) and some don't.

## Motivation

Some sequences do not converge, but they unambiguously tend to $+\infty$ or $-\infty$. For instance consider the sequences $a_{n}=n$, $b_{n}=2^{n}$ and $c_{n}=-n+2\cdot (-1)^{n}$:
* The sequence $a_{n}=n$ tends towards $+\infty$.
* The sequence $b_{n}=2^{n}$ tends towards $+\infty$ even faster.
* The sequence $c_{n}=-n+2\cdot (-1)^{n}$ tends towards $-\infty$.
We will give a mathematical definition which classifies these sequences as diverging towards $+\infty$ or $-\infty$. 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 $+\infty$, not to $-\infty$ .
The sequence $e_{n}=(-1)^{n}\cdot n$ is unbounded, but contains parts (subsequences) tending towards $+\infty$ and parts tending towards $-\infty$ :
* 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 $+\infty$ nor to $-\infty$.

## Definition

We have observed some sequences, which tend towards $+\infty$ or $-\infty$ . How can we give a mathematically precise classification for this observation?
Let us start with divergence to $+\infty$: "A sequence diverges to $+\infty$" means that it grows larger than any number $S$, no matter how large (since $+\infty$ 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 $+\infty$:
Definition (Divergence to $+\infty$)
A sequence $(a_{n})_{n\in \mathbb {N} }$ diverges to $+\infty$, if for any $S\in \mathbb {R}$ almost all sequence elements are greater or equal to $S$ . That means, for all $S$ , there is an index $N$, such that $a_{n}\geq S$ for all $n\geq N$. In quantifier notation:
$\forall S\in \mathbb {R} \,\exists N\in \mathbb {N} \,\forall n\geq N:a_{n}\geq S$
We can translate this quantifier notation piece by piece:
${\begin{array}{l}\underbrace {{\underset}\forall S\in \mathbb {R} } _{{\text{For any real }}S}\ \underbrace {{\underset}\exists N\in \mathbb {N} } _{{\text{ there is a minimal index }}N,}\ \underbrace {{\underset}\forall n\geq N:} _{{\text{such that for all indices }}n\geq N}\\[0.5em]\quad \quad \underbrace {{\underset}a_{n}\geq S} _{{\text{ the element }}a_{n}{\text{ is bigger or equal }}S}\end{array}}$
Divergence to $-\infty$ is just defined analogously:
Definition (Divergence to $-\infty$)
A sequence $(a_{n})_{n\in \mathbb {N} }$ diverges to $-\infty$, if for any $s\in \mathbb {R}$ almost all sequence elements are lower or equal to $s$ . That means, for all $s$ , there is an index $N$, such that $a_{n}\leq S$ for all $n\geq N$. In quantifier notation:
$\forall s\in \mathbb {R} \,\exists N\in \mathbb {N} \,\forall n\geq N:a_{n}\leq s$

## Notation

For $(a_{n})_{n\in \mathbb {N} }$ diverging to $+\infty$ , we write
$\lim _{n\to \infty }a_{n}=+\infty$
Analogously, for $(a_{n})_{n\in \mathbb {N} }$ diverging to $-\infty$ , we write
$\lim _{n\to \infty }a_{n}=-\infty$

## Examples

Example
${\begin{aligned}\lim _{n\to \infty }n&=+\infty \\[0.5em]\lim _{n\to \infty }n^{2}&=+\infty \\[0.5em]\lim _{n\to \infty }-n&=-\infty \end{aligned}}$
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, $\lim _{n\to \infty }q^{n}=\infty$ for $q>1$. However, if $q\leq -1$, there is neither a divergence to $+\infty$ nor to $-\infty$: 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 $q\leq -1$ . For $|q|<1$ and $q=1$, it is convergent.

## Improper convergence

The notation $\lim _{n\to \infty }a_{n}=+\infty$ 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 $+\infty$ is not a number. However, the divergence to $\pm \infty$ has a lot in common with convergence to a number (except for boundedness):
| Convergence | Divergence to $\pm \infty$ |
| In each $\epsilon$-neighbourhood (interval), we can find almost all sequence elements. | In each interval $[S,\infty )$, we can find almost all sequence elements. |
| All subsequences converge to the same limit. | All subsequences also diverge to $\pm \infty$. |
| Every convergent sequence is bounded. | Every sequence diverging to $\pm \infty$ is unbounded. |
Especially, some of the limit theorems hold true for $\pm \infty$ and in some cases, one may threat a sequence diverging to $\pm \infty$, similar as a convergent sequence. Sometimes, the divergence to $\pm \infty$ is even called "improper convergence". However, always keep in mind that an 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 $\lim _{n\to \infty }(a_{n}\cdot b_{n})=\lim _{n\to \infty }a_{n}\cdot \lim _{n\to \infty }b_{n}$ in case both sequences diverge. If one sequence converges and a second one diverges, some rules may also turn out wrong. For instance,
$1=\lim _{n\to \infty }\left(n\cdot {\frac {1}{n}}\right)=\lim _{n\to \infty }n\cdot \lim _{n\to \infty }{\frac {1}{n}}=\infty \cdot 0=0$
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! In the article "Divergence to infinity: rules" we will find out, which rules for limit calculations still hold for improperly convergent sequences.