For an infinite sequence of real numbers that monotonically grows, is infinity considered a limit?

Question

My question is on the topic of "infinity" as a limit.

Does a sequence like $1,2,3,4,5,6,7,\ldots$ have a limit of infinity, or is it considered as not having a limit. If infinity is considered a limit, is this sequence considered convergent then?

My reason for this question is to further distinct these two sequences: $1,2,1,3,1,4,1,5,1,6,1,\ldots$ and $1,2,4,3,7,4,10,5,\ldots$

As per my last question from a previous post

My professor stated that the first sequence is considered as "not going to infinity", while the second one does.

He hasn't given a formal definition of what that means, but I'm going to assume that means $\lim a_n = \infty$ and $\lim a_n$ does not exist.

If infinity is considered a limit point, would that make the second sequence convergent, converging to infinity, while the first sequence as being divergent, with the limit not existing?

In that case, wouldn't this clash with some definitions? In a wikipedia article, it says

"Zaporedje $2, 4, 6, 8, 10,\cdots$ je divergentno in ima nepravo stekališče neskončno"

which in translation means

"The sequence $2, 4, 6, 8,\cdots$ is divergent and it has the false limit point of infinity"

But as previously stated, this form of sequence would have indeed a limit, of infinity, and thus would be convergent?

Answer 1

This is arguably a matter of convention and definition, but - a priori - no.

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Short Version:

If a sequence $\newcommand{\N}{\mathbb{N}} \newcommand{\R}{\mathbb{R}} \newcommand{\seq}[1]{\left( #1_n \right)_{n \in \N}} (a_n)_{n \in \N}$ grows without bound (i.e. $a_n \to \pm \infty$), we say that

$$\lim_{n \to \infty} a_n = \pm \infty$$

not as a statement that the limit exists or that the sequence converges, but as an encoding of how it diverges. It diverges, but in a special way that we can encode in familiar notation easily, but diverges nonetheless.

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Long Version:

This is a matter of definition insofar as "what does it mean to converge?"

We consider a metric space $(X,d)$. ($X$ is just some set, and $d$ is a function which takes in two inputs, and outputs the distance between them. This may not match the distance you and I are familiar with in Euclidean space.)

A metric function must satisfy four axioms: for any $x,y,z \in X$,

• $d(x,y) \in \R_{\ge 0}$
• $d(x,y) = 0$ if and only if $x=y$
• Symmetry: $d(x,y) = d(y,x)$
• Triangle Inequality: $d(x,z) \le d(x,y) + d(y,z)$

In such a space, a sequence $\seq x$ in $X$ is said to converge to $x \in X$ if

$$(\forall \varepsilon > 0)(\exists N \in \N)(\forall n \ge N) \Big( d(x_n,x) < \varepsilon \Big)$$

or, less formally,

$$\lim_{n \to \infty} d(x_n,x) = 0$$

The key point here is that the limit is in the space itself.

So, if we want to speak about convergence to $\infty$ in $\R$, we really need to speak about the extended reals, $\overline{\R} := \R \cup \{+\infty,-\infty\}$.

But now you should see why, if $d(x,y) := |x-y|$ is the usual distance in $\R$, if you extend it to the usual arithmetic with $\infty$, that $(\overline{\R},d)$ is actually not a metric space, so the notion of convergence there (and thus "convergence to infinity") is not really sensible. (It's as simple as realizing, for instance, that $d(x,\infty) = \infty \not \in \R_{\ge 0}$ for any $x \in \R$. Hence $d$ is not a metric.)

(Some more discussion here.)

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Addressing a Potential Follow-Up:

A reasonable conclusion to go to from here would be "okay, well, can we speak about convergence to infinity under a different metric?" (Though I suppose you could loosen the restrictions on finite distances, e.g. like here.)

Any given set can be given some sort of distance function, albeit it may be limited to a trivial one, such as

$$d(x,y) := \begin{cases} 1 & x \ne y \\ 0 & x = y \end{cases}$$

A metric for $\overline{\R}$, per Ross Milikan's answer here would be

$$d(x,y) := |f(x)-f(y)|$$

where

$$f : \overline{\R} \to \left[ - \frac \pi 2, \frac \pi 2 \right] \text{ has } f(x) := \begin{cases} \pi/2 & x = +\infty \\ -\pi/2 & x = -\infty \\ \arctan(x) & x \in \R \end{cases}$$

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What About Limit Points?

You also brought up the idea of "would $\infty$ be a limit point?"

To recall: for a topological space (e.g. metric space) $X$, a point $x \in X$ is a limit point of a subset $S \subseteq X$ if, given any open neighborhood $U_x$ of $x$, we have $U_x \setminus \{x\} \cap S \ne \emptyset$.

The key point to think about here is that we need to speak about an overarching set/subset/superset structure. In particular, each in the chain must, essentially, inherit the topological structure from those above it.

Hence, $\infty$ needs to be a point in something, so it seems natural to start regarding $\R$ as a subset of $\overline{\R}$.

However we run into the same snag as before, in that $\overline{\R}$ is not a metric under the usual distance function for $\R$.

Now, of course, we have seen the existence of a metric for $\overline{\R}$. If we restrict that to $\R$, that too will be a metric, so we can speak of convergence in the general sense I opened with.

So, in the resulting topology, would $\infty$ be a limit point of $\R$?

It is enough to look at open balls in metric spaces, so take $B_r(\infty)$ for $r>0$. (A ball, in $\overline{\R}$, centered at $\infty$ with radius $r$.) Formally,

$$B_r(\infty) = \left\{x \in \overline{\R} \, \middle| \, d(x,\infty) < r \right\}$$

Of course,

$$d(x,\infty) = \left| \frac \pi 2 - \arctan(x) \right|$$

for $x \in \R$, and $d(-\infty,\infty) = \pi$. Then if $r > \pi$, $B_r(\infty) = \overline{\R}$. For lesser $r$, you will get some interval $(a,\infty]$ (all reals from some $a \in \R$ onward, and including $\infty$) -- this is easy to see from the limiting behavior of $\arctan(x)$, and I leave formal justification to you.

Either way, in this limited sense, sure, $\infty$ is a limit point of $\R$. Just under a different metric so that $\overline{\R}$ can even be spoke of as a metric space.

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Your Two Sequences:

Something you'll learn about is the formal definitions for infinite limits. I've bombarded you with enough as is, and other posts have addressed this point in particular.

However something you'll learn is that, if a sequence converges to some limit, every subsequence must do so too. This is why your first sequence does not converge and why we would not even say $a_n \to \infty$, since every other term is $1$. The sequence as a whole is not going to infinity, it keeps bouncing back down. The other sequence, overall, is bouncy but still trends upward.

You can chase these around with formal definitions if you want.

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Summary:

• Under the Euclidean metric, we only say a sequence "goes to infinity" to give a little detail as to how it diverges. (That is, sequences $a_n$ where $a_n \to \infty$ are just a special subset of divergent sequences.)
• Such a sequence can be said not to converge, formally, because the Euclidean metric does not extend to $\overline{\R}$.
• We can introduce a metric on $\overline{\R}$, however. In that sense, we can speak of $\infty$ as a limit point of $\R$.

Answer 2

There are two common ways of defining what $$\lim_{n \to \infty} a_n = \pm\infty$$ means for a real-valued sequence $a_n \in \mathbb R$.

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The first definition is commonly encountered in high school and first-year undergraduate textbooks, where "divergence to infinity" is defined separately from normal convergence to a limit.

That is, in addition to the normal definition of convergence where, for a sequence $a_n \in \mathbb R$ and a (possible) real limit $x \in \mathbb R$, $$\lim_{n \to \infty} a_n = x \overset{\rm\ def.}{\iff} \forall \epsilon > 0\ \exists N_\epsilon \in \mathbb N\ \forall n \in \mathbb N: (n \ge N_\epsilon \implies |a_n - x| \le \epsilon)$$ there are also two additional definitions: $$\lim_{n \to \infty} a_n = +\infty \overset{\rm\ def.}{\iff} \forall M \in \mathbb R\ \exists N_\epsilon \in \mathbb N\ \forall n \in \mathbb N: (n \ge N_\epsilon \implies a_n \ge M) \\ \lim_{n \to \infty} a_n = -\infty \overset{\rm\ def.}{\iff} \forall M \in \mathbb R\ \exists N_\epsilon \in \mathbb N\ \forall n \in \mathbb N: (n \ge N_\epsilon \implies a_n \le M)$$

The symbols $+\infty$ and $-\infty$ in these definitions are not defined to mean anything outside the formal expressions defined above; in particular, they are not considered to be real numbers, which are the only kinds of numbers normally considered in textbooks using these definitions. Also, sequences satisfying one of the latter two definitions are not considered convergent, but rather are said to diverge to (positive or negative) infinity.

The main advantage of this definition is that it avoids using anything but real numbers — the symbols $+\infty$ and $-\infty$ are treated as just notational devices, not as actual numbers. It also doesn't require any conceptual background beyond what's normally introduced in advanced high school math. The down side, of course, is that we need separate terms and definitions for different cases of something that — intuitively and notationally — seems like it ought to be one and the same thing.

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The second definition is more commonly found in graduate and advanced undergraduate textbooks that either introduce or assume the student to be familiar with the general notion of topological convergence of sequences:

Definition: A sequence $a_n$ in a topological space $X$ is said to converge to a limit $x \in X$ (written $\lim_{n \to \infty} a_n = x$) if and only if, for any open neighborhood $B \subset X$ of $x$, there exists an $N_B \in \mathbb N$ such that, for all $n \in \mathbb N$, $n \ge N_B \implies a_n \in B$.

(Some textbooks may instead directly define topological convergence for nets and then simply consider a sequence to be a particular kind of net. This yields a more general and arguably more elegant definition, but also a more abstract one.)

Given this definition, we may then define the extended real number line $\overline{\mathbb R} = \mathbb R \cup \{-\infty, +\infty\}$, equip it with the "natural" total order where $-\infty < x < +\infty$ for all $x \in \mathbb R \subset \overline{\mathbb R}$ (and real numbers retain their normal order) and the associated order topology, and then simply interpret $\lim_{n \to \infty} a_n = \pm\infty$ for real-valued sequences $a_n$ as indicating convergence in $\overline{\mathbb R}$.

This approach has the advantage that we no longer need multiple separate definitions: the same definition of (topological) convergence now works for all limits in $\overline{\mathbb R}$. It also generalizes readily, matching the definition of convergence in other metric and topological spaces that the student is likely to be familiar with at this stage. The down side, of course, is that it requires the student to know at least a little bit of general topology, which tends to be considered an advanced topic.

(Technically, since $\overline{\mathbb R}$ is metrizable, it's possible to do all this using only the concept of convergence in a metric space, which is commonly introduced before general topology. The awkward bit here is that there's no way to equip $\overline{\mathbb R}$ with a metric that would both generate the standard order topology on it and agree with the usual euclidean metric $d(x,y) = |x-y|$ on $\mathbb R$.)

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Ps. A somewhat common compromise by intermediate-level textbook authors is to define convergence of sequences using the first type of definition, with only real numbers, but also briefly mention the existence of the extended real number line $\overline{\mathbb R}$ and the fact that, using more advanced mathematical concepts, it's possible to unify the separate definitions under a single definition of convergence in $\overline{\mathbb R}$. After all, merely defining $\overline{\mathbb R}$ as an ordered set does not require any particularly advanced background knowledge, and the general idea is quite simple and intuitive. It's only when rigorously defining the order topology and convergence under it that things get complicated.

Answer 3

It really is all a matter of semantics and definitions. I cannot a priori say $n\to\infty$ without first defining what that means. The usual definition is

We say that $\lim_{n\to\infty}x_n=\infty$ if for all $M>0$ there exists $N\in\mathbb{N}$ such that $n\geq N$ implies $x_n>M$.

Do these 'limits' behave in the same way as regular limits. In some ways yes, in other ways no. For example, suppose $a_n\to A$ and $b_n\to B\neq 0$. Then

$$a_n+b_n\to A+B$$

$$\frac{a_n}{b_n}\to \frac{A}{B}$$

For infinite limits, suppose $a_n\to\infty$ and $b_n\to \infty$. Then

$$a_n+b_n\to\infty$$

but $\frac{a_n}{b_n}$ could go to $0$, $\infty$, or really any positive real number.

Answer 4

The usual definition of $a_n\to\infty$ is that for all $M$ (equivalently: $M>0$) there exists $N\in\mathbb{N}$ such that for all $n\geq N$ we have $a_n \geq M$. Notice the second sequence in your question satisfies the definition above, while the first sequence does not.

Intuitively: for any given bound, the sequence will eventually go past it (and never go below it again).

Whether or not one considers a sequence that satisfies this definition as a "convergent" sequence or not is a matter of definition. One might say that it "converges to infinity", another might insist that it "diverges to infinity". It all depends on what terminology you introduced in your course. Given your question it seems that you've defined that such sequence "converges to infinity" (some would define that it "converges to infinity" but that it is not a "convergent sequence").

Since you mentioned monotonicity, notice however that a sequence does not have to be monotone in order to satisfy this definition. For instance, the sequence $1,10,2,20,3,30,\ldots$ is not monotone but does converge to infinity.

Answer 5

In resolving your confusion, it will be helpful to realize that we often work with two different types of limits in analysis literature, with one defined on the usual real number line, $\mathbb{R}$, and the other on the extended real number line, $\overline{\mathbb{R}} = \mathbb{R}\cup\{-\infty, +\infty\}$.

Since both types of limits are extremely useful, one needs to come up with some terminology that can distinguish them. Unfortunately, there seems to be no universally accepted convention. In one extreme case, one simply rejects the idea of $\pm\infty$ being limits, whereas in the other extreme, one not only regards $\pm\infty$ as viable limits but also considers the sequences tending to $\pm\infty$ to be convergent. In most cases, however, people choose somewhat in the middle of these two extremes. My impression is that the following choice of terminology is gaining popularity in the literature:

• There is no arguing with the notion of "$(a_n)$ converging to a real number".

• A sequence $(a_n)$ of real numbers is said to have a limit if (i) $(a_n)$ converges to a real number, or (ii) $(a_n)$ diverges to $+\infty$, or (iii) $(a_n)$ diverges to $-\infty$. Equivalently, $(a_n)$ is said to have a limit if it has a limit in the extended real number line $\overline{\mathbb{R}}$.

• So, in this convention, "having a limit" is a strictly more general notion for a sequence than "being convergent".

Of course, this terminology may not be what your professor is using. So, you may want to ask your professor to clarify this.

Answer 6

There are many great answers here, but I would simplistically point out that having terms go to infinity literally means said sequence is UNLIMITED!

So, why would you want to declare such an unlimited sequence as limited or convergent? Doing so wreaks havoc on the notion of a limit existing!