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Consider a divergent series that tends to infinity such as $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots$. The limit of this series is unbounded, and I have often seen people say that the sum 'equals infinity' as a shorthand for this. However, is it acceptable to write $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots = \infty$ in formal mathematics, or is it better to denote that the limit is equal to infinity? If so, how does one do this?

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    $\begingroup$ Infinity in limiting analysis is really just a notation. So yes, you can. $\endgroup$ – Don Thousand Nov 29 at 23:00
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    $\begingroup$ In formal mathematics you would just say the sum diverges. You would not say that it equals infinity. You might want to say that the partial sums are unbounded - that's one way a sum can diverge; there are others. $\endgroup$ – Ethan Bolker Nov 29 at 23:01
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    $\begingroup$ To everyone that tells you "this is just a notation" ask them what isn't a notation. Yes, it is as valid as writing $1+1=2$. It is actually better than saying "diverges", since it actually gives more information. Not all divergent series diverge by tending to infinity. $\endgroup$ – conditionalMethod Nov 29 at 23:19
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    $\begingroup$ It's perfectly valid if working in the extended reals. $+\infty$ is a legitimate value in that case as long as we adhere to the restriction that $\infty - \infty$ is undefined and $0 \times \infty$ may or may not be defined depending on context. Here there's no issue as we're summing a series of nonnegative values. Lebesgue integration and measure theory would actually be more cumbersome if we didn't admit $\infty$ as an allowable value. $\endgroup$ – Bungo Nov 29 at 23:21
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    $\begingroup$ One unambiguous thing to say is that the series $1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots$ "diverges to $+\infty$". This has the advantage of reminding one of convergence to (a finite value) while simultaneously reminding one that the series does not, in fact, converge --- instead, it diverges. I've seen this terminology in some textbooks, for example Fitzpatrick's "Advanced Calculus" (which I quite like). $\endgroup$ – Lee Mosher Dec 2 at 0:04
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A plea from a high-school teacher: Do not use $=\infty$ anywhere in any school course. The idea of infinity already causes a great deal of confusion amongst schoolchildren (as well as fascination), and our best approach is to say firmly that, 'Infinity is not a number, but an idea, and the notation $\infty$ is used as shorthand in a few standard pieces of notation.'

Yes, I know that everything in mathematics is an idea, and that every mark on the page is notation, but leave that to university philosophy classes. What we don't want is monstrosities such as $\infty - \infty = 0$ and $\infty / \infty = 1$, which school-children routinely arrive at if we give even a hint that infinity is a number.

(I wouldn't be using $=\infty$ in undergraduate courses either. Without a rigorous axiomatic understanding of mathematics, plus some projective geometry, it is misunderstood and misused.)

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  • $\begingroup$ Thank you. Man, I was starting to feel a bit alone here. $\endgroup$ – Arnaud Mortier Dec 4 at 15:54
  • $\begingroup$ @Sid Thanks, I think this is the most appropriate answer considering the level at which I am working at. Can you clarify the following statements to ensure I am not getting confused: a) there are different number systems, including the real numbers, and the extended reals, b) calculus tends to use the former, whereas analysis the later, c) No matter which number system you are using, series are defined to equal the limit of the partial sums, d) Since infinity is not treated like a number in calculus, you cannot say the limit "equals infinity" (rather, there is no limit) $\endgroup$ – Joe Dec 4 at 18:26
  • $\begingroup$ @Sid e) In analysis, where the concept of infinity is rigorously defined, it is acceptable to say the sum "equals infinity" because the limit can be said to be infinity (as infinity can in this particular case be treated as a number) $\endgroup$ – Joe Dec 4 at 18:28
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Yes - it is both very common and entirely correct to do so. There is a bit of formal trickery here because $\infty$ is not a number, but you can do analysis with it anyways - meaning limits and that sort of thing. In particular, there is a set called the affinely extended reals which is basically the real numbers $\mathbb R$ along with two new objects $\infty$ and $-\infty$, one at each 'end'. This is a topological space, meaning that you can take limits in it, but be careful that some things like $∞-∞$, $0·∞$, $0/0$ and $∞/∞$ are undefined.

Consider that, for real numbers $x$, the definition of a sequence $s_n$ converging to $x$ is as follows:

For any $\varepsilon >0$, there exists some $N$ such that if $n>N$ then $|s_n-x|<\varepsilon$.

This can be rewritten as saying:

For any open interval $I$ containing $x$, there exists some $N$ such that for all $n > N$ we have $s_n\in I$.

The idea behind either of these definition is that if we choose some "neighborhood" of $x$ - consisting of $x$ and at least some positive radius around $x$ - the sequence eventually is constrained in that neighborhood. More formally, a neighborhood of a real number is any set $S$ containing an open interval around $x$. Then, you can define convergence to $x$ as follows:

For any neighborhood $I$ of $x$, there exists some $N$ such that for all $n >N $ we have $s_n\in I$.

To define limits to $\infty$ and $-\infty$, one just needs to define their neighborhoods. In particular, $\infty$ is meant to be the "upper end" of the real line - and being close to $\infty$ means that a number is very large. So one defines a neighborhood of $\infty$ to be any set $I$ containing an interval of the form $(C,\infty]$ for some $C\in\mathbb R$. Then, we say

$\lim_{n\rightarrow\infty} s_n = \infty$ if for every neighborhood $I$ of $\infty$, there exists some $N$ such that if $n>N$ then $s_n\in I$.

This is equivalent to saying that $s_n$ converges to $\infty$ if, for every $C$, there exists an $N$ such that if $n>N$ then $s_n > C$ - which is the usual definition you find in textbooks (but note that it is actually a theorem - a consequence of the definition of $\infty$!) - and that in any context that you might allow a statement like $\lim_{n\rightarrow\infty}s_n = \infty$, you might as well be working in the extended reals.

Then, since infinite sums are just limits of partial sums, it is perfectly rigorous to write $$\sum_{n\rightarrow\infty}\frac{1}n = \infty$$ and to know that this truly means that the left hand side evaluates to $\infty$, not to think that this is some special statement where equality is not equality. This is actually very common in real analysis (the branch of mathematics dealing with limits, continuity, differentiability, and all that stuff) - especially in subfields like measure theory and sometimes in the theory of metric spaces as well.

However, it is also important to know that many people do not share the view that $\infty$ is always a perfectly valid object, defined by its neighborhoods. So, even though you would technically be right to write such an equality, it might not go over well with your audience nonetheless - and you should keep your audience in mind whenever you write anything because "formal correctness" is no substitute for "understood by your audience" - and you will often encounter examples of things which are technically correct, but might confuse or annoy your audience nonetheless.

(Sidenote: The limit $n\rightarrow\infty$ in the subscript $\lim_{n\rightarrow\infty}s_n$, as you might notice, is also defined by the neighborhoods: we get to restrict $s_n$ by forcing $n$ to lie in some neighborhood of $\infty$ that we get to choose - which is what's going on when we say that there's some $N$ so that if $n>N$, blah blah blah)

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    $\begingroup$ Although everything in your answer is obviously correct, I wanted to avoid this point of view because it's going to induce more confusion among weak students than it's going to help those who understand well already. But that's just me adapting my speech to my audience :-) $\endgroup$ – Arnaud Mortier Nov 30 at 8:34
  • $\begingroup$ Hello! I clarified the "extended reals", added a link to the wikipedia article, and added a tiny warning to be careful with the affinely extended reals. I hope you don't mind. =) $\endgroup$ – user21820 Nov 30 at 16:14
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    $\begingroup$ @ArnaudMortier It depends on what is being taught. In calculus, yeah, extended reals are only marginally helpful and likely to be very confusing. Beyond introductory analysis, they start to fit easily into the theory that's already being taught and introducing them unifies a lot of edge cases involving $\infty$ into conceptual theorems - and the alternative is having lots of edge cases whose proofs are almost identical to the general case (except for switching out neighborhoods!), which imposes a real mental load on students, and lets errors creep in through these edge cases. $\endgroup$ – Milo Brandt Nov 30 at 21:42
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    $\begingroup$ @lalala That's not really the case - in every metric space, all convergent sequences are Cauchy. The trouble is that the usual metric on $\mathbb R$ doesn't extend to the extended reals (...even if you allow infinite distances, since then you get the wrong topology) - so you can't speak of Cauchy sequences at all in it. (Of course, the extended reals are metrizable since you can map them bi-continuously to $[-\pi,\pi]$ by $\tan^{-1}$, but then you're introducing arbitrary choice in the metric and also, then it's true that all convergent sequences are Cauchy) $\endgroup$ – Milo Brandt Nov 30 at 21:51
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    $\begingroup$ @MiloBrandt For sure, yet there is nothing in the way the question is being asked, that could make you think that the OP is anywhere beyond elementary matter. $\endgroup$ – Arnaud Mortier Dec 1 at 10:04
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The only thing that you can correctly write is that a limit is equal to infinity.

Be aware that writing this is nothing but a shortcut for the longer sentence $\forall M\in \Bbb R,\exists n\in \Bbb N, $ etc. It doesn't mean that there is a limit which is a number and that this number is $\infty$.

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    $\begingroup$ This is the best answer in my opinion. As long as we are talking about the standard integers and the standard reals, the sum doesn't exist in this case, so the equation is just a shorthand for a more formal limit expression. $\endgroup$ – Cuspy Code Nov 30 at 10:49
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  • Let $\{s_n\}$ be a sequence of real numbers with the following property:

    For every real $M$ there is an integer $N$ such that $n\geq N$ implies $s_n\geq M$.

    We then write $\displaystyle \lim_{n\to\infty}s_n=\infty. $ This expression has a clear definition as above. We sometimes write $+\infty$ instead of $\infty$.

  • In your example, by the definition above, one writes $\displaystyle \sum_{k=1}^\infty\frac{1}{k}:=\lim_{n\to\infty}\sum_{k=1}^n\frac{1}{k}=\infty. $

  • In real analysis, one very often works with the extended real numbers, especially in the theory of measure and integration. One can interpret the symbol $\infty$ as an extended real number and the expression $\displaystyle\lim_{n\to\infty}s_n=\infty$ can be understood as convergence in $\mathbb{R}\cup\{\infty,-\infty\}$ with the order topology. One may also work with the extended non-negative real axis $[0,+\infty]$ with the extended topology. See for instance, this set of notes.

  • Moreover, observe that a function $n \mapsto x_n$ from the extended natural numbers ${\Bbb N} \cup \{\infty\}$ (with the order topology) into a topological space $X$ is continuous if and only if $x_n \to x_{\infty}$ as $n \to \infty$, so one can interpret convergence of sequences as a special case of continuity.


"The limit of this series is unbounded"

Notes: "bounded/unbounded" is a concept for some subset of real numbers. One can say "some sequence of real numbers is unbounded", or "certain subset of real numbers is unbounded". It does not make sense to say that the limit of [the partial sums of] a series is "unbounded". In your particular case, one could say that the limit of [the partial sums of] the series "is not a real number" or "is not finite".

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It is widely accepted for divergent series that tends to infinity the following short notation

$$\sum_{n=1}^\infty \frac1n =\infty$$

instead of the extended version usually introduced at first by the formal and rigorous definition

$$\lim_{N\to \infty }\sum_{n=1}^N \frac1n =\infty,\quad\sum_{n=1}^N \frac1n \to \infty \iff \forall M\in \mathbb R \quad\exists N_0\in \mathbb N \quad\forall N\ge N_0 \quad \sum_{n=1}^N \frac1n>M$$

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    $\begingroup$ I really don't like the first, even if it is used, first because the OP specifically asked about formal mathematics, and second because it is not good practice. Students write plenty of incorrect mathematics with no need for us to help them. $\endgroup$ – Arnaud Mortier Nov 29 at 23:07
  • $\begingroup$ @ArnaudMortier I specify better that is is just a "short notation". $\endgroup$ – user Nov 29 at 23:09
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    $\begingroup$ Note to anyone: $=\infty$ is not normally used as a notation for divergent series. The series $\sum_{n=1}^{\infty}(-1)^n$ diverges in the Cauchy sense, but $\sum_{n=1}^{\infty}(-1)^n=\infty$ isn't used to denote this. $\endgroup$ – conditionalMethod Nov 29 at 23:25
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    $\begingroup$ @conditionalMethod I mean "divergent to $\infty$" of course since we are dealing with the harmonic series. BTW I would say that $\sum_{n=1}^{\infty}(-1)^n$ is "not convergent" and not that it diverges. $\endgroup$ – user Nov 29 at 23:28
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    $\begingroup$ If there is a series that deserves to be called divergent is $\sum_{n=1}^{\infty}(-1)^n$. Divergent = dis (apart) + vergere (to bend, turn, tend toward) = go in different directions. $\endgroup$ – conditionalMethod Nov 29 at 23:37

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