divergent vs undefined integrals/summations I have tagged this as a soft questions because it is more a question of language/notation but for certain summations/integrals these terms seem to be used interchangably and I would like some clarity on their actual meanings.
As a first example I have the sum:
$$S=\lim_{n\to\infty}\sum_{n=0}^n(-1)^n=1-1+1-1...$$
If we stop this at a given point (as long as the first value is taken as 1) this can either take the value of $1$ or $0$, both of which are finite values, but I have all kinds of manipulations of this series and there are ways of making this series take really any value you choose. This suggests that its value is undefined, but is it finite or would it be considered convergent?
Another example is the integral:
$$I=\lim_{n\to\infty}\int_0^n\sin(x)dx$$
obviously it doesn't matter particularly if this is cosine or sine or any other periodic wave satisfying the following conditions:
$$f(x+T)=f(x)$$
$$\int_a^{a+T}f(x)dx=0$$
but in a similar way to the first function we can break these integrals off at any point and it will have a finite value:
$$I(n)=\int_0^n \sin(x)dx=\int_0^{2\pi k}\sin(x)dx+\int_{2\pi k}^n\sin(x)dx$$
where $2\pi k<n<2\pi(k+1)$ and $k\in\mathbb{N}$. Clearly this first integral is equal to $0$ and at most this second integral can equal $1$ so it seems fair to say that:
$$|I(n)|\le1$$
for any finite $n$. However what happens when $n\to\infty$?
 A: A sequence that does not converge is said to diverge. This is the definition of divergent sequence. So asking, “does this sequence diverge?” is automatically answered when you answer the question, “does this sequence converge?”. Your first series does not converge, i.e. it diverges, so your proposed number $S$ does not exist.
Recall that a series is just a sequence of partial sums, and the order of terms in a series does matter unless there are either finitely many positive or finitely many negative terms.
Your first series is known as Grandi’s Series, by the way.
I always find it useful to think of a series as a limit of a sequence of partial sums, which is discussed in the introduction to the Wikipedia entry on series. So what the series $$1-1+1-1...$$ really is, is the limit if it exists, as $n \to \infty$ of the sequence (of partial sums): $(1,0,1,0,1,0,...)$. Clearly this sequence, which "is basically" your series, does not converge, i.e. it diverges.
Note that the above series is not the same as $$-1+1-1+1-...,$$ whose corresponding sequence of partial sums is $(-1,0,-1,0,-1,0,...)$, which is different to the previous sequence. [It is true that this series/sequence of partial sums diverges because again, the sequence of partial sums does not converge to anything.]
In other words, in a series, the order of terms matters.
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Furthermore, there is in fact a theorem called the Riemann Series Theorem, which states that for any number $L$, the terms of a conditionally convergent series can be rearranged so that the series is equal to $L$. For example, this says that you can rearrange the terms in the sum of the alternating harmonic series to make the sum equal to any number, including very large negative or very large positive numbers. Personally I find this fact remarkable. This is proven in Rudin's PMA Theorem $3.54$.
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And whilst Grandi’s Series does not converge in the usual sense with partial sums, there are non-standard summation methods under which it converges: See: https://en.m.wikipedia.org/wiki/Summation_of_Grandi%27s_series
It is clear that the Cesàro Sum of Grandi’s series converges to $\frac12$ and you could also look into the Abel Sum and other methods of summing the series if you want...
Now to answer some of your questions:
"...but I have all kinds of manipulations of this series and there are ways of making this series take really any value you choose." This is false: no Grandi's series does not converge, and in fact it is easy to see that no rearrangement of Grandi's series converges.
"This suggests that its value is undefined, but is it finite or would it be considered convergent?" I think the term you're looking for is "bounded", and yes, Grandi's series is bounded. Obviously, some rearrangements of Grandi's series will be unbounded, e.g. $-1+1+1-1+1+1-1+1+1-1...$ will tend to $+ \infty$, thus being an unbounded sequence of partial sums.
I haven't spent much time thinking about the second part of your question: the sine/cosine stuff, but I get the feeling that my answer can be applied to that as well... in fact, it is answered by: https://en.wikipedia.org/wiki/Integral_test_for_convergence , which says that a series converges $ \iff$ it's corresponding integral converges, so long as the series and the integral have the same limits. So really, it's the contrapositive to this theorem that answers the second half of your question.
A: If using hyperreal numbers, there is a way to do the summation so that you get an actual value for the divergent series (though often it is a specific hyperreal infinite value).  The paper (coauthored by me) is "Hyperreal Numbers for Infinite Divergent Series".
The reason why you can regroup it and get different answers is that, in a hyperreal sense, you are actually changing the number of items being summed.  That is, if $\omega$ is the unit infinity, having $\omega - 1$ terms or $\frac{\omega}{2}$ terms is a different number of terms than the original.
The easiest way to find a value, that we found (see Section 9), is to take a discrete integral (i.e., a symbolic summation), and then replace all occurrences of $-1^\omega$ with zero.  This is not proven in the paper, but we have an unpublished proof of it, and the paper cites a proof that $\sin(\omega) = 0$ in the surreal numbers.
In this case, the sum is $\frac{1}{2}$, which matches the Cesaro summation.
In short, using hyperreal numbers allows you to do things that you couldn't do when you restrict yourself to the reals.
