$$\frac{n}{\infty} + \frac{n}{\infty} +\dots = \frac{\infty}{\infty}$$

You can always break up $\infty/\infty$ into the left hand side, where n is an arbitrary number. However, on the left hand side $\frac{n}{\infty}$ is always equal to $0$. Thus $\frac{\infty}{\infty}$ should always equal $0$.

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    so $\lim\limits_{x\to\infty} \frac{10x}{x}=0$? – The Count Dec 04 '16 at 19:24
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    This is a sincere question, that shows some real thought, so it doesn't deserve the downvotes. – littleO Dec 04 '16 at 19:25
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    Infinity is not a number. – Joffan Dec 04 '16 at 19:25
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    I have changed my mind and upvoted. Every time I try to think of a one-sentence refutation of your idea, I end up unsatisfied. – The Count Dec 04 '16 at 19:30
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    See [Extended number line : Arithmetic operations](https://en.wikipedia.org/wiki/Extended_real_number_line#Arithmetic_operations) : "The arithmetic operations of $\mathbb {R}$ can be **partially** extended to $\mathbb {R} \cup \{ \infty \}$ as follows:... " Emphasis added on **partially**; on the LHS the ratio $\dfrac n {\infty}$ is defined (and thus meaningful) in the extended domain but an infinite sum is **not** defined in "usual" arithmetic and thus we cannot even try to extend it. – Mauro ALLEGRANZA Dec 04 '16 at 20:16
  • lim(x->infinity) x^2/x != 0. – Joshua Dec 04 '16 at 22:31
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    Why do you think that $\infty / \infty$ could be broken up into the left hand side anyways? –  Dec 04 '16 at 22:48
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    You are assuming that multiplication will stay continuous if the quotient of infinities is defined. It does not stay continuous. – Henricus V. Dec 04 '16 at 22:58
  • @Hurkyl I don't see how it can't – JobHunter69 Dec 04 '16 at 23:11
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    @Goldname Well, it's not even clear what the left side means: how do I calculate infinite sums? And once I've fixed a way to do so, how do I justify that those expressions are equal? There are some [pretty weird properties](https://en.wikipedia.org/wiki/Riemann_series_theorem) of infinite sums of *real numbers*, let alone stuff involving infinity . . . – Noah Schweber Dec 05 '16 at 00:11
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    Or I could point out that divisor and dividend are the same, therefore it equals one. And it still would be nonsense. – WGroleau Dec 05 '16 at 06:10
  • @TheCount Well if you follow my reasoning it does equal to 0, but I'm trying to figure out what is wrong: $lim_{x\to\infty}\frac{10x}{x} = \frac{10 + 10...}{1+1...} = \frac{10}{1+1...}+\frac{10}{1+1...}...=0$ – JobHunter69 Dec 05 '16 at 06:18
  • @Goldname: Of course I can't. I'm pointing out that my example makes just as much sense as the OP's, i.e., none. – WGroleau Dec 05 '16 at 06:24
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    @goldname: $lim_{x\to\infty}\frac{10x}{x}$ has a very specific definition and $\frac{10 + 10...}{1+1...}$ does not. Neither $10+10\dots$ nor $1+1\dots$ are real numbers. You need to define them and show that your definition does not lead to any contradictions with, for example, addition and multiplication. Until you do, $\frac{10 + 10...}{1+1...} = \frac{10}{1+1...}+\frac{10}{1+1..}$ is just pretty nonsense. "I don't see how it can't" is not very good mathematics. – Steven Alexis Gregory Dec 05 '16 at 08:26
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    Some ∞ are bigger than others. Some ∞ we know for certain are twice the size of some other ∞ (hence in some cases ∞/∞ = 2 makes perfect sense). As such ∞/∞ can technically be anything. – slebetman Dec 05 '16 at 09:13
  • See https://en.wikipedia.org/wiki/Indeterminate_form for L'Hôpital's rule for evaluating the indeterminate forms 0/0 and ∞/∞, developed in 1694. In order to provide a relevant answer, Why are you asking this question in the latter part of 2016? – Jules Bartow Dec 05 '16 at 16:07
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    @JulesBartow To see what is wrong with the wrong way. – JobHunter69 Dec 05 '16 at 16:11
  • I think part of the issue is that "$n$ as an arbitrary number" repeats an infinite amount of times. In essence, what you have is $\frac{n \times \infty}{\infty}$ – Gil Keidar Feb 02 '18 at 00:35

9 Answers9


How do you know $0+0+0+0+...... = 0$? If you think about it $0 + 0 + 0 +..... = 0\times\infty = 0\times 1/0 = 1$. (or any other number). We clearly are dealing with a value on which standard arithmetic doesn't apply. (Hence "infinity is not a number".)

So the question becomes what does apply and how do we deal with this? And that is not an easy/simple question. It's not hard... but it's not simple. Bottom line, finite rules of arithmetic do not always apply, and such instincts lead to common traps.

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    I liked this answer the best, the 0*1/0 hits the Crux of the matter – Sidharth Ghoshal Dec 04 '16 at 19:37
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    The problem with this answer is that $0 + 0 + \ldots = 0$ is actually true. –  Dec 04 '16 at 22:23
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    @Hurkyl I don't see the problem. It may be true, but the answer shows that one should be careful when trying to apply familiar arithmetic rules to infinite sums. – This site has become a dump. Dec 04 '16 at 22:26
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    @servaes precisely. It's not enough that 0+0+0+ .... = 0. We must know *why*. Afterall $0 + 0 + 0+.... = 0 *\infty = 0*1/0 = 1$ so ... one of those basic statements is either wrong or doesn't mean what we think it does. If $1/\infty = 0$ which.... is something people say.... then $1/infty + 1/infty ... = 0+ 0+.....$ which ... is not so clear. – fleablood Dec 04 '16 at 22:47
  • Is that "because" there in the 2nd paragraph supposed to be "becomes"? – Kyle Kanos Dec 05 '16 at 00:58
  • Very much concise, clear, and short enough not to get lost in. – Simply Beautiful Art Dec 05 '16 at 01:30
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    I think this answer lacks somewhat some emphasize on the fact that undefined concepts are arbitrary. You say it "(or any other number)" but it is not that clear, IMHO. You could say "If you think about it, we can consider that $0+0+\dots = \lim_{k\to\infty}\sum_{n=1}^{k}0 = 0$ or we can also consider that this is $0\times \infty=0/0$, which is undefined and thus to which we can assign any value." – MoebiusCorzer Dec 05 '16 at 11:11
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    Yes, I personally would "take" 0+0+0... to mean $\lim \sum 0$ and I would take $0*\infty = \lim 0*n = \lim \sum^n 0$. But I'd take $k/\infty = \lim k/n = 0$ and ... is equality not transitive. $\lim \sum 0 = \lim \sum \lim k/n = ???$ Point is, I am making assumptions and I can't assume "well, *this* step is obviously right". All must be questioned. – fleablood Dec 05 '16 at 17:38
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    @fleablood, in regard to your comment to Morgan Rodgers, I am confused why it isn't obvious that 0 + 0 + .... = 0? I think since each term is 0 their sum will always be 0, no matter how many terms we add. Don't we do similar stuff in limits when if some function f(x)'s value grows without bounds at some x=a, we say the limit of f(x) at a is infinity? – HeWhoMustBeNamed Sep 09 '18 at 17:04
  • We spend an entire lecture and frequently half a chapter defining precisely *what* the definition of a lim is and when we can and can not determine whether sequences converge and if you search this site there are *hundreds* of question of how to prove $\lim_{x\to 2} x^2 = 4$ or what do I figure what the $\lim_{n\to \infty} \frac {x^2 + 5}{2x^3 - 27}$ is. If you put the same care in *defining* what $0 + 0+ .... $ an infinite times then sure. But unless you *have* put that time in, I will not listen to any "well, it's obvious..." – fleablood Sep 09 '18 at 18:42
  • " I am confused why it isn't obvious that 0 + 0 + .... = 0? I think since each term is 0 their sum will always be 0, no matter how many terms we add." Replace "$0+0+... = 0$" with "$3.1415926.....$ is rational" and replace "each term is 0" with "each decimal point gives a rational number", you will get the following FALSE argument with the EXACT same reasoning: I do not see why it isn't obvious that $3.141526.... = \pi$ is rational. After all $3$, $3.1$, $3.14$ etc. are all rational no matter how many decimal places we take it to. What is the difference between your argument and this one? – fleablood Sep 09 '18 at 18:47
  • Or: Isn't it obvious that $0.999999....$ is less than $1$. Each term takes us closer to one but we are always less than $1$ no matter how many terms you add. So isn't it obvious that $0.999999.....$ is less than $1$. – fleablood Sep 09 '18 at 18:49
  • And finally. Consider what is the length $[0,1]$? It is $1$. What is the lenght of $[0,3]$ it is $3$. And $[0,1]\cup [1,3]$? Is is $1 + 2=3$. And what is the length of $\{1\}$? It is $0$. It is a single point. It has no length so it is $0$. So what is the length of $\{0\}\cup \{.1\} \cup \{3\}$. Each has a length of $0$ so it has length $0+0+0$. Okay what if we did this for ever single real number in $[0,1]$. Each number has a length of $0$. So length of $[0,1] = 0+0+0+0+.....= 0$ But the length of $[0,1]$ is $1$. – fleablood Sep 09 '18 at 18:55
  • @ΛRYΛN that was my point. The OP is claiming it's "obvious" that $0+ 0+0+....$ will always be equal to $0$ no matter how many times you add it. I claim unless it is precisely defined what adding $0+0+ ....$ an infinite number of times *means* and prove it *is* zero there is nothing obvious about it. As *AN ANALOGY* if we don't define or use any rigor we may make the *EXACT* same argument that it's "obvious" $0.99...$ for any finite number of $9$s is less than $1$ so $0.999...$ with infinite $9$s is "obviously" less than $1$. Which is bullshit. I was pointing out why the argument is false. – fleablood Dec 20 '20 at 17:35

The existing answers are very good; let me give yet another one.

Consider the counter-argument that ${\infty\over\infty}=\infty$: We have $\infty=2^\infty$, so $${\infty\over\infty}={\infty\over 2^\infty}={\infty\over 2}\cdot {1\over 2}\cdot {1\over 2}\cdot {1\over 2}\cdot...$$ But ${\infty\over 2}$ is infinity, and stays infinity no matter how many times we divide it by $2$. So ${\infty\over \infty}=\infty$.

This is exactly the same "shape" as the argument you give, but yields the opposite answer; so something must be wrong with this kind of argument.

The problem lies in the "number" $\infty\over\infty$. It hasn't been precisely defined. Now, some of the time in math it's clear what we mean when we write some complicated expression; however, this isn't one of those times. We need to sit down and precisely define what this thing is.

When we try to do so, we'll find something surprising: it doesn't follow the usual laws of arithmetic! For example, consider the following "proof" that $1=2$: $${1\cdot{\infty\over\infty}}={\infty\over\infty}={2\infty\over \infty}=2\cdot{\infty\over\infty},\mbox{ so }1=2.$$ Since $1\not=2$, at least one of the steps there has to be nonsense. The obvious candidate is the claim that we can cancel $\infty\over\infty$. This suggests that ${\infty\over\infty}=0$; however, then we have $\infty\cdot {1\over\infty}\not=1$, which ruins one of the basic properties of division!

It gets worse: in both arguments, we need to perform infinitely many operations (either infinitely many sums, or infinitely many products), so we need to define how those work. At first glance it seems like limits can help us there, but again, we'll find that there's no good way to define the infinite operations we need which matches with our intuition.

This all points to the following fact:

In order to make sense of arithmetic involving infinity or infinite operations, we need to make some choices in how we precisely define the various operations on infinity; and no matter how we make these choices, some "obvious" properties of numbers will fail to hold.

Picture trying to fit a carpet into a room that's too small. You can maybe get lots of it to lie flat, but somewhere it's going to fold over, or stick up, or crumple.

Noah Schweber
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    On the subject of making sense of infinite operations, you may be interested in various kinds of infinite summation that let us sum divergent series. For example, do you think $1-1+1-1+1-1+...={1\over 2}$? Then [Cesaro summation](https://en.wikipedia.org/wiki/Ces%C3%A0ro_summation) might be right for you! – Noah Schweber Dec 04 '16 at 20:07
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    @SimpleArt No, that kind of regrouping isn't allowed in Cesaro summation - see the wikipedia page. Basically, any method for summing divergent series is going to have to be "fragile". – Noah Schweber Dec 05 '16 at 02:03
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    Yes, I know, but many of those viewing this post do not. I was pointing out that perhaps you might want to reconsider leaving that comment there. – Simply Beautiful Art Dec 05 '16 at 02:04
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    @SimpleArt I disagree. The point of that comment was precisely that there are many ways to sum a non-convergent series. I think it *does* make sense in this context, because the ambiguity of infinite operations is indeed one of the problems with the OP's argument. I am leaving the comment. – Noah Schweber Dec 05 '16 at 02:24
  • Fine, as you wish. I just think summing divergent series is completely different from the OP's context, but I am not the judge of that. Cheers anyways for the nice answer. (Also deleted comments that would probably confused others) – Simply Beautiful Art Dec 05 '16 at 14:20

The problem is that before you can even consider a proof of a statement, it has to be clear what that statement means. I have no idea what $\infty/\infty=0$ could possibly mean.

The original answer ended here.

Since not everyone was happy with that answer, let me elaborate. Since $\infty$ is not a number, I just do not know what $\infty/\infty$ should mean. Now you could respond that you are using formulas like $\infty+\infty=\infty$ and have been told that that one is true. Well, it is a useful short-hand, but what people really mean when they write this is the following:

If $(a_n)$ and $(b_n)$ are sequences of real numbers such that $\lim_{n\to\infty} a_n=+\infty$ and $\lim_{n\to\infty} b_n=+\infty$, then $\lim_{n\to\infty} a_n+b_n=+\infty$.

Now this is a perfectly fine statement, and it happens to be also true. Another perfectly fine statement (but a wrong one) would be:

If $(a_n)$ and $(b_n)$ are sequences of real numbers such that $\lim_{n\to\infty} a_n=+\infty$ and $\lim_{n\to\infty} b_n=+\infty$, then $\lim_{n\to\infty} a_n/b_n=0$.

Now maybe one could understand $\infty/\infty=0$ in this way, but you would have to say so. And now you also see that it would be hard to see how the “proof” in this question relates to this. Maybe one could make the following weaker statement:

If $(b_n)$ be a sequence of real numbers such that $\lim_{n\to\infty} b_n=+\infty$, then $\lim_{n\to\infty} n/b_n=0$.

One could then say that the “proof” in your question is to be understood as follows: \begin{multline}\lim_{n\to\infty} \frac{n}{b_n} =\lim_{n\to\infty}\underbrace{\left(\frac1{b_n}+\cdots+\frac1{b_n}\right)}_{\text{$n$ summands}} =\\=\lim_{n\to\infty}\underbrace{\left(\lim_{n\to\infty} \frac1{b_n}+\cdots+\lim_{n\to\infty}\frac1{b_n}\right)}_{\text{$n$ summands}} =\\=\lim_{n\to\infty}\underbrace{(0+\cdots+0)}_{\text{$n$ summands}}=\lim_{n\to\infty}0=0. \end{multline} Now here at least every expression has a defined meaning and we can ask: Where is this wrong? Well, it turns out that the second equality is wrong, and the reason why is again boring: There is just no reason that this equality should hold. It is not enough that it somehow looks nice, we would have to cite some theorem that we have proved earlier to justify this equality, and there is none. (Of course we can pinpoint the wrong equation by plugging in a counterexample. For example for $b_n$=n you can evaluate each expression and get $1=1=0=0=0=0$.)

Anyway, my point is that before we can ask why something is wrong it first has to make minimal sense. And the proof in your question is, as they say, “not even wrong”, because it is unclear what the statement is.

Carsten S
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    I find this answer unsatisfying. I get what is being said and I agree with it, but it means exactly what the OP says it means: dividing an arbitrarily large number by itself is 0. Makes no sense, but it is what we are working with as a hypothesis. – The Count Dec 04 '16 at 20:05
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    I think this answer as "satisfying". You say " exactly what the OP says it means: dividing an arbitrarily large number by itself is 0" but I don't so what "an arbitrarily large number" is supposed to be. Infinity isn't *an* arbitrarily large number, is it? – fleablood Dec 04 '16 at 21:53
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    When I write $\infty + \infty = \infty$, I really mean to do arithmetic on extended real numbers. –  Dec 04 '16 at 23:25
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    @Hurkyl, sure, one can also mean that, but then one has to say that one means that, and one has to say what the definitions are, unless they are clear. – Carsten S Dec 04 '16 at 23:27

First of all $\infty$ is not a real number, so it's unclear what you mean by $n/\infty$ in the first place.

But, let's suppose that you extend the real numbers by introducing a new number "$\infty$" which has some expected properties, such as $n/\infty = 0$ for any real number $n$. You would still need to define $\infty/\infty$ before you can refer to it, otherwise we don't even know what "$\infty/\infty$" means. You could define it to be $0$ if you liked, but somebody else could equally well argue that $\infty/\infty$ should be defined to be equal to $10$ (for example) because $$ \lim_{x \to \infty} \frac{10x}{x} = 10. $$ Another person could argue that $\infty/\infty$ should be defined to be equal to $\infty$, because $$ \lim_{x\to\infty} \frac{x^2}{x} = \infty. $$ There is no value that we could assign to $\infty/\infty$ that stands out as being more valid than any other possible value. So, we simply leave $\infty/\infty$ undefined.

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  • Comments are not for extended discussion; this conversation has been [moved to chat](http://chat.stackexchange.com/rooms/49663/discussion-on-answer-by-littleo-whats-wrong-with-this-reasoning-that-infinity). – Daniel Fischer Dec 06 '16 at 14:33

If $\frac\infty\infty=0$, then we should have $\infty\cdot 0 = \infty$.

But OP's main equation $\frac{n}{\infty}+\frac{n}{\infty}+\frac{n}{\infty}+\cdots=\frac\infty\infty$ (which is supposed to equal $0$) suggests that $\infty\cdot 0 =0$.

How can $\infty\cdot 0$ equal both $0$ and $\infty$?

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  • Isn't multiplying infinities undefined? I tried to see how I can simplify the infinity/infinity problem by breaking it apart without breaking any rules. – JobHunter69 Dec 05 '16 at 03:21
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    I don't see why (in this hypothetical situation) we can't look at OP's equation as saying infinity times 0 is $\frac\infty\infty$, since we're adding $0$ (in the form of $\frac{n}{\infty}$) to itself infinitely many times and getting a result of $\frac\infty\infty$. – paw88789 Dec 05 '16 at 03:24
  • Yes, OP is indicating with his equation that 0 = inf/inf, but OP is not indicating that inf*0=inf, because to get inf on rhs, you must first say inf/inf = 1. In other words, aren't you saying that (inf / inf) * inf = 0 * inf. And then inf/inf = 1 to finally get inf *0 = inf? – JobHunter69 Dec 05 '16 at 03:28
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    So we no longer require $\frac{a}{b}=c$ to be equivalent to $bc=a$? This principle seems important to me if one wants to have a meaningful conversation about division. – paw88789 Dec 05 '16 at 03:36
  • Well I think we must require that only if b/b = 1. But currently, our result indicates that infinity/infinity = 0 (also in actuality infinity/infinity is undefined, so it can't be 1) – JobHunter69 Dec 05 '16 at 03:38
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    @paw88789: This shows that the OP's arithmetic leads to a contradiction, but it doesn't enlighten him at all, because it doesn't explain *why* that contradiction arises (i.e. the deep reasons that cause it). To me, this post is not an answer. – Alex M. Dec 05 '16 at 11:18
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    @Goldname, I find funny the way you refer to yourself as "the OP", instead of just "me". – Alex M. Dec 05 '16 at 11:19

When you learned how to extend arithmetic from the natural numbers to the integers to the rational numbers to the real numbers and to the complex numbers, one of the main motivations was to preserve the laws of arithmetic.

The situation with the extended real numbers (i.e. the real numbers along with $\pm \infty$) is different — the goal is not to preserve the laws of arithmetic, the goal is to have continuity.

Consequently, arguments that involve naively applying the ordinary laws arithmetic to extended real numbers are quite unreliable.

This is compounded by the fact the argument involves infinite summation. Infinite sums can fail to satisfy many of the laws that finite sums obey, so that's a second source of unreliability.


I think an issue with your reasoning as well is that to "split up infinity" you would have to have an infinite sum of finite terms. You can see with Riemann integration that an infinite sum of 0s is not necessarily 0. Of course none of this is rigorous but I think it should help you intuition a little.

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Infinity is a fine number (and so is its negative counterpart), and is surprisingly well-behaved if you unpack it carefully.

Without going into too much detail about what that means, just note that Cantor showed that there are many types of infinities with different characteristics.

This means that we have to be clear about what sort of infinity we're talking about.

Most of the existing answers implicitly interpret infinity as some number implied to be the result of some unbounded number of arithmetic accumulations, which makes sense given the framing of your question. So what kind of infinity is reached in that fashion? $\omega$!

$\omega$ is a number that, informally, is always bigger than any real number you can produce using finite arithmetic steps. It's a member of the hyperreals, symbolized as $\mathbb{R}^*$. The nice thing about the hyperreals is, anything you can state in first-order logic (i.e., no tricks with arguments on sets) about the reals is still true. This is known as the transfer principle.

In the hyperreals are also infinitesimals $\epsilon$, which are the reciprocals of the infinitudes. They work just as you'd expect: a real number divided by an infinitude is an infinitesimal. So the hyperreal number line looks like this (Wikipedia):

hyperreal number line

With these tools, we note that each of the terms in your series are infinitesimal:

$$\frac{n}{\infty} = \epsilon$$

You have an infinitude of them, so the series can be rewritten as:

$$\epsilon + \epsilon + \dots = \omega\epsilon$$

This is an indeterminate form in the hyperreals (see chapter 1 of Keisler), just as you would expect. The reason for this is because $\frac{n}{\infty}$ is not zero, and in the first place, it wouldn't make sense for a infinitely divisible real number to be zero in that case.

So you can have infinity and infinitesimals as your equation depicted, but you still have to acknowledge indeterminates if you do so.

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    $\infty$ is greater than every real number, and consequently it's greater than every hyperreal number too (in particular, $\omega \neq \infty$, no matter which unlimited number you choose for $\omega$). Also, there is a difference between an infinite sum and a (hyper)finite sum of $\omega$ terms. Also, this topic is unrelated to Cantor's work on cardinality and ordinal numbers. –  Dec 04 '16 at 23:19
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    This isn't really accurate; $\omega$ is **not** an element of the hyperreals. It *is* an element of the *surreals*, but that's a very different object. In particular, your answer suggests that there is only one kind of infinity in the hyperreals, and similarly only one kind of infinitesimal. This is false; e.g. if $\epsilon$ is infinitesimal, $\epsilon^2$ is "even more" infinitesimal, and this is crucial to the structure of the hyperreals. Indeed, there is *no* distinguished infinite/infinitesimal element of the hyperreals. – Noah Schweber Dec 04 '16 at 23:23
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    The hyperreals are a perfectly valid framework for analysis, but they're more complicated than you're answer suggests; meanwhile, the ordinals (Cantor's transfinites), where $\omega$ lives, are a much worse setting for arithmetic. The surreals are somewhat a mixture of the two, but transfer issues there are much more complicated; they lack the reflexive ease of the hyperreals. – Noah Schweber Dec 04 '16 at 23:26
  • @NoahSchweber We're not arguing about notation, are we? I haven't studied the surreals, and I was just using $\omega$ since the Wikipedia article from which the image is sourced uses it. Keisler uses $H$ and $K$ for hyperreal infinites. – bright-star Dec 04 '16 at 23:26
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    @bright-star Agh, I didn't know that - that's awful. I'm slightly arguing over notation, just because of how well-established the set-theoretic meaning of "$\omega$" is, and how easily it is to get confused when thinking about ordinals and hyperreals at the same time. I didn't realize wikipedia conflated the two; my apologies. My main point, though, is about the hyperreals being more complicated than is suggested by that picture (again: dang you, wikipedia!). – Noah Schweber Dec 04 '16 at 23:29
  • @Hurkyl The OP didn't define what sort of $\infty$ they were interested in, so I made the same attempt at inference as every other answerer. Also, "greater than every real number" does not *per se* imply "greater than every hyperreal number"--is there something else you meant there? Also, The Cantor stuff is about sets and cardinality, yes, but the refined intuition that you gain from his discovery is illustrative of the variety of infinitudes. – bright-star Dec 04 '16 at 23:29
  • @bright-star: Your introduction of $\omega$ sounds like you are computing a limit ordinal. (I kind of thought so too but wasn't sure, which is why I made a couple of remarks on $\omega$ that should trigger a reaction if you really meant it) –  Dec 04 '16 at 23:29
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    @bright-star: Transfer principle: all standard concepts, including the extended real numbers $\pm \infty$ (or whatever hacks one prefers to use them as notation without actually using them as mathematical objects) transfer and apply equally well to the nonstandard model. –  Dec 04 '16 at 23:31
  • @NoahSchweber I agree. Keisler wrote one entry-level calculus book using NSA, but he has a graduate-level text that actually goes into the basis of that first book. Still, that entry-level text (linked in answer) has a similar image in its first chapter to help the reader get a feeling for the numbers. I wouldn't accept it uncritically either (it's just the level of the question didn't seem to match a longer digression). – bright-star Dec 04 '16 at 23:31
  • @Hurkyl You're right, I didn't mean it that way. Any editing you have in mind? – bright-star Dec 04 '16 at 23:32
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    Personally, I like pictures. Also, this isn't necessarily a bad answer, as it refines the question into "what do you mean by '$\infty$'?" – Simply Beautiful Art Dec 05 '16 at 01:43
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    -1 for "fine number", and for mentioning Cantor without distinguishing between ordinals and cardinals. – Martin Argerami Dec 05 '16 at 14:10
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    @bright-star: It’s good that you linked to [the Wikipedia article](https://en.wikipedia.org/wiki/Hyperreal_number) in your answer, and mentioned it in a comment.  But, as I understand [Stack Exchange policy](//math.stackexchange.com/help/referencing) (and IANAL), you’re expected to visibly state the source of any content that you have copied (and I’m referring specifically to the number line picture).  Unless, of course, you are [M.Romero Schmidtke](http://enciclopedia.us.es/index.php/Usuario:M.Romero_Schmidtke), in which case (a) never mind, and (b) good work! – Scott Dec 05 '16 at 19:19
  • @Scott Named the source. Sorry, I'm used to thinking of URLs as citations in themselves at this point. – bright-star Dec 05 '16 at 19:58

You cannot use an "infinite" distributive law with division by the symbol $\infty$. In other words the statement $$\frac{\sum_{n=1}^\infty a_n}{\infty} = \sum_{n=1}^\infty \frac{a_n}{\infty}$$ is not valid if $\sum_n a_n$ is a divergent series. For the left-hand side cannot have any meaning when that sequence diverges, whereas the right-hand side is the zero series (provided that each $a_n$ is an ordinary finite number) which has the sum $0$ for any reasonable summation definition.

The law $\frac{a+b}{d}=\frac{a}{d}+\frac{b}{d}$ is true for real or complex numbers $a$, $b$ and $d$ provided $d\ne 0$, but it does not quite generalize to the form you use.

Jeppe Stig Nielsen
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