Are the integers closed under addition... really? 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 $S_1$.
$S_2 = \frac{1}{4}$
 A: 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.
A: 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.
A: 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 $S(-1)$. 
A: 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.
A: 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!
A: There's no need for philosophy here.
What you call $S$ isn't a sum at all, it is the limit of a sequence, (if it happens to even converge).
A: $$1-1+1-1+\cdots=\frac{1}{2}$$ is called Cesaro summabilty. It is an averaging used in Fourier series like the Heaviside step function, but it is not a sum in any usual or convergent sense. Also, when we use series, make sure that they are used within the radius of convergence, otherwise strange and unexpected results will occur. 
The Platonic idea of math as an eternal verity of the Form World is being challenged by people who want to "use" math, bending it to their purposes. This isn't always bad, I guess, since we do have non-Euclidean geometries, calculus, etc. I think we have to be careful that this bending of math doesn't lead us into temptation. Exercise caution...
