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Okay so I'm a 3rd year undergraduate studying Mathematics. I've proved in group theory countless times that the integers are closed under addition.

It's obvious to me that they are.

However this has just put a little bit of a spanner in the works and I'd like to see your thoughts on this.

Take the infinite sum: $S = 1 + x + x^2 + x^3 +... = \frac{1}{1-x}$ for $x<1$.
[The geometric series is only convergent for $|x| < 1$. Maybe this is part of the issue. ---PLC]

[There are many other cases of these, take the well known one from physics that $\sum_{n\in \mathbb{Z}}^\inf n = 0 + 1 + 2 + 3+.. = \frac{-1}{12}$]
A well known result.

Now let's differentiate this, with respect to $x$.

$\frac{d S}{dx} = 1 + 2x + 3x^2 + 4x^3+ ...= \frac{1}{(x-1)^2}$

Now treating $S$ as a function of $x$ let's substitute in $x = -1$

This gives $S = 1 - 2 + 3 - 4 + ... = \frac{1}{4}$

Now I have a sum of integers added together and I get: $\frac{1}{4}\notin \mathbb{Z}$

Thus the integers are not closed under addition.

Now I assume the problem here is as always: infinity. The gap between infinite sums and normal sums always seems to provide these little strange problems. It's commonly accepted that both $S = \frac{1}{4}$ and the integers are closed under addition.

Let's get as philosophical here as you like..

EDITS..

My motivation for this question came from a Numberphile video on youtube: http://youtu.be/PCu_BNNI5x4
http://www.youtube.com/watch?v=w-I6XTVZXww
Who state an alternate proof as:

Given $S_1 = 1 +1 - 1 - 1.. = \frac{1}{2}$. Proved in the first link.
Let $S_2 = 1 - 2 + 3 - 4 + ... $

So $2S_2 = 1 - 2 + 3 - 4 + ...$
$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + 1 - 2 + 3 - 4 + ...$

Now by adding coloums we can easily see that:
$2S_2 = 1 - 1 + 1 - 1 + .. = \frac{1}{2}$ by $S_1$.
$S_2 = \frac{1}{4}$

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31  
What you call "addition" is actually divergent series regularization via analytic continuation (and as you observe, $\Bbb Z$ isn't closed under this process). Whether you call this process "addition" or not depends on how literary and poetic you want to wax. What is your actual question? –  anon Jan 10 at 22:21
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The geometric sum formula you mention works only for $|x|<1$, not $x<1$. –  JiK Jan 10 at 22:23
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«Now I assume the problem here is as always: infinity.» Hm, not really: The problem here is that you are using concepts outside of their field of validity. In an abelian group there is no notion of «infinite sum» so anything you find using it is going to be absurd. –  Mariano Suárez-Alvarez Jan 10 at 22:28
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I think the main problem here is that you are equivocating: using "closed under addition" to mean two different things. If you fix this problem, does a substantial question remain? –  Trevor Wilson Jan 10 at 22:34
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You have a typo in the line after Now let's differentiate this, with respect to x. you need a minus sign on the RHS –  Nasser Jan 10 at 22:35

6 Answers 6

up vote 56 down vote accepted

When it is said that "$X$ is closed under binary operation $\circ$", it means that for any $a$ and $b$ in $X$, $a \circ b$ is in $X$. It is easy to prove (by a simple induction) that any finite sum is therefore closed in $X$.

However, infinite sums are defined with a limit (of the partial sums), which means they don't just depend on the operation $\circ$, but also require a topological structure defined on $X$. Now the integers $\mathbb{Z}$ do have a standard topological structure in addition to their algebraic structure, it's the discrete topology, and it comes from the order on $\mathbb{Z}$. However, in this system, there is actually no limit of the sequence of partial sums $1$, $1 - 2$, $1 - 2 + 3$, ... (*) and so no infinite sum. In fact, an infinite sum of integers can only have a limit if all but finitely many of its terms are $0$. Another subtle flaw is that when you took a "derivative", that means you passed from $\mathbb{Z}$ to $\mathbb{R}$, and evaluated a function on $\mathbb{R}$ on the right side, to obtain a "sum" for the left (which may be a valid technique, giving a form of "divergent summation", but it's important to remember this is actually a generalization beyond the usual and strict "limit" definition of infinite summation). So you left the integers, and thus it is no surprise you'd get a non-integer result. The important point to remember though is that infinite sums and finite sums are not the same thing -- one is purely algebraic, the other leverages additional (topological) structure on the set in question, and closure is strictly algebraic.

Finally, it's important to note that this series doesn't have a sum in $\mathbb{R}$ either in the strict, limit sense.

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Summation is defined to be finite. Infinite series are just limit of a sequence which is defined as partial sums.

The integers, if so, are closed under finite sums. And therefore by definition, closed under summation.

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8  
@jakey You're mistaking natural language meaning with mathematics. –  Git Gud Jan 10 at 22:27
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@jakey It's true that sometimes people talk about "infinite sums" or the "sum of an infinite series" but in any case the point is that infinite sums and finite sums are different things, so $\mathbb{Z}$ can be closed under one but not under the other. –  Trevor Wilson Jan 10 at 22:27
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@jakey: In addition to the comments by Git Gud and Trevor Wilson, let me ask you, are the rational numbers closed under summation? In fact, are the real numbers closed under summation? It's easy, in almost every structure, to find a series which is divergent to infinity. Closure under summation means that finite summations are still within the structure. Infinities are generally handled by limits and topological structures, which are additional to the basic group structure (needed for defining the summation in the first place). –  Asaf Karagila Jan 10 at 22:29
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Asaf I agree, I like that answer! You are very correct and I'm glad you aren't taking as negative route as others against me! I shouldn't be punished for merely asking! –  jakey Jan 10 at 22:31
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I guess I also find it weird that you and Git Gud each deny that $\sum_{n=1}^{\infty} a_n$ is a "sum". Yes, it is the limit of a sequence of partial sums. We call that type of limit the sum of the series, right? It's not the same thing as a finite sum and is not defined in exactly the same contexts, but it's certainly defined in this context, right? (The concept is defined, I mean.) (Git Gud's answer confuses me more; it seems to express doubt on whether the series converges; but surely he knows it diverges. I felt like I was missing something on reading these answers...) –  Pete L. Clark Jan 11 at 10:43

The series representation $$ 1+z+z^2+\ldots=\frac{1}{1-z} $$ is only valid for $|z|<1$ (in the sense that the left-hand side converges only in that region of the complex plane). So you can't assert this equality (or any of its derivatives) at $z=-1$: the sum $1-1+1-1+\ldots \neq 1/2$, at least not without an agreed-upon convention for the meaning or value of a non-convergent series. The problem doesn't have to do with infinity in this case, and the closure of $\mathbb{Z}$ is safe: the sum of an infinite series of integers will be an integer if the sum converges at all. (Of course, it can only converge if there are only finitely many nonzero terms.) In other cases, though, a group that is closed under finite summation may well not be closed under "infinite summation" (e.g., $\mathbb{Q}$ doesn't contain all of its limit points); you could reasonably say that "infinity is the problem" there.

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6  
Actually, I find somewhat elegant the fact that applying the formula to 1+2+4+8+etc. yields -1. That value would be the length of an infinitely long binary number that was all "1"'s, which is what one would receive if one subtracted 1 from 0. –  supercat Jan 11 at 0:04
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@Shaun: your first comment is incorrect: the series converges exactly where mhqxxx says it does. The series is the Taylor series of a function which is holomorphic on a larger domain, but that's not the same statement at all. –  Pete L. Clark Jan 11 at 4:58
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@Shaun: The principle of analytic continuation applies to holomorphic functions. What you've written on the left hand side is not a function at all for $|z| \geq 1$ since it's given by a divergent series. So the identity is not "valid" for exactly this reason: to be a valid identity each side has to mean something! –  Pete L. Clark Jan 11 at 10:17
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(Here is what I think you meant to say: your identity holds for all $|z| < 1$, and the right hand side is defined and holomorphic for all $z \neq 1$. Therefore the right hand side is the -- unique, by the principle of analytic continuation -- extension of the left hand side to the larger domain. So if you want to give a meaning to the left hand side for $|z| \geq 1$ but unequal to $1$, it is reasonable from the perspective of complex function theory to have it mean $\frac{1}{1-z}$.) –  Pete L. Clark Jan 11 at 10:21
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@supercat: You're on to something! In fact the series literally does converge to $-1$ with respect to the $2$-adic norm. This a is a norm in which large powers of $2$ are close to $0$. –  Pete L. Clark Jan 11 at 10:50

There's no need for philosophy here.

What you call a $S$ isn't a sum at all, it is the limit of a sequence, (if it happens to even converge).

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To clarify the situation:

  • The integers are closed under addition. Any finite sum of integers is an integer.
  • The integers are also complete under the usual metric. If an infinite series of integers converges in this metric, it must converge to an integer.
  • The series $1-2+3-4+\cdots$ does not converge; its "sum" does not exist.
  • The Abel sum of the series $1-2+3-4+\cdots$ is $\frac14$. This example proves that the Abel sum of a series of integers is not necessarily an integer.

There are no contradictions here!

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thank you! I'm impressed with how quickly this so so many views and answers. It is all so clear now! –  jakey Jan 10 at 22:43
    
well I'm glad I've made you think about it! I consider that a result! –  jakey Jan 11 at 12:30

This seems a bit circular. You took a derivative of a power series and then tried to argue that the sum of a set of integers is not an integer. Why should it be here? You're calculating $S^\prime(-1)$, not $S(-1)$.

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The sum of a set of integers should be an integer because $\mathbb{Z}$ is a closed subset of $\mathbb{R}$. The fact that the suggested answer is $\frac{1}{4}$ indicates that this is a divergent series for which a sum is being attached via some other summability method (e.g. Abel summation). –  Pete L. Clark Jan 11 at 5:17
    
You might have a point. However, in the concrete example it is obvious that after differentiation and setting $x=-1$ all terms of the power series are still integers, so the real difficulty is elsewhere. (It is an interesting question whether one can come up with a integer-term convergent power series at $x_0\in\Bbb C$, which if instead taken at a variable $x$, differentiation with respect to $x$, and again setting $x=x_0$ in the derivative gives a series converging to a non-integer value. I guess not, because integer-term series only converge if the terms ultimately become $0$ forever.) –  Marc van Leeuwen Jan 11 at 13:39

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