Why can infinite series be summed different ways to get different results? $$S = 1 - \frac12 + \frac13 - \frac14 + \frac15 - \frac16 + \frac17 - \frac18 + \frac19 - \frac1{10} + \frac1{11} - \frac1{12}\ldots\text{(to infinity)}$$
Rearranged, this series looks like:
$$S = \left(1 - \frac12\right) - \left(\frac14\right) + \left(\frac13 - \frac16\right) - \left(\frac18\right) + \left(\frac15 - \frac1{10}\right) - \left(\frac1{12}\right) + \left(\frac17 - \frac1{14}\right) \ldots\text{(to infinity)}\\
 S = \left(\frac12\right) - \left(\frac14\right) + \left(\frac16\right) - \left(\frac18\right) + \left(\frac1{10}\right) - \left(\frac1{12}\right) + \left(\frac1{14}\right) \ldots\text{(to infinity)}$$
This rearranged infinite series contains every number that the original infinite series had.
Further,
$$ 2S = 1 - \frac12 + \frac13 - \frac14 + \frac15 - \frac16 + \frac17 - \frac18 + \frac19 - \frac1{10} + \frac1{11} - \frac1{12} \ldots\text{(to infinity)}$$
Thus: $2S = S$
$2 = 1$
Mathematics disproven.  Sorry.
Jokes aside, I know that infinite series can be calculated in different ways to get different results.  My question is: Why?  While it makes sense with Grandi's and similar series, it doesn't make sense to me for a series whose final term is $\frac1\infty = 0$.
 A: This is a good opportunity to point out something that many people don't get, or at least get accustomed to glossing over. When you write something like $S = 1 - 1/2 + 1/3 - 1/4 + 1/5$, you really are saying that $S$ is a sum; you can explicitly add the terms (and get $47/60$ in this specific case).
On the other hand when you write something like $S = 1 - 1/2 + 1/3 - 1/4 + 1/5 - \cdots$, this is a completely different animal. You can't explicitly add the terms because there are infinitely many of them. $S$ is not a sum, even though it is traditional to say so (I call it a sum as well). It may be spoken of as an "infinite sum", and it's convenient to regard it that way be cause in the right conditions it acts like a sum, but it is not--it is actually a limit.
For each $n$, the partial sum $S_n = 1 - 1/2 + \cdots + 1/n$ is indeed a sum which can be computed by explicitly adding the terms. The "infinite sum" $S$ is not a sum at all, but rather the limit $\lim\limits_{n\rightarrow\infty}S_n$, if it exists.
Often it is fine to forget about that and pretend $S$ is really a sum, but that can get you in trouble sometimes. So always keep in the back of your mind the notion that infinite sums are really limits of finite sums.
A: If a series 
$$\sum_{i=0}^\infty a_i$$
is absolutely convergent, i.e. the sum $$\sum_{i=0}^\infty |a_i|$$ is convergent, then it can be shown that for any permutation (bijective function) of natural numbers $\pi$, the sum
$$\sum_{i=0}^\infty a_{\pi(i)}$$ is the same.
For a series which does not absolutely converge, the same proof does not work. In fact, if the series is conditionaly convergent (it converges, but not absolutely), you can prove that you can choose ANY real number and can find a permutation $\pi$ such chat the sum will converge toward that number. The way I understand is that for non-absolutely-convergent series, the sum is "just barely well defined". It's only "truly" well defined for absolutely convergent series.
A: To get a formal argument you can look at a proof of the Riemann rearrangement theorem - I'll try to give a more intuitive explanation (although the proof I know of the theorem is based on the same idea).
Essentially, this happens when both the positive and negative terms of the series can become arbitrarily small, but the positive terms still sum to $+\infty$ and the negative ones to $-\infty$. A standard example of this is the alternating harmonic series $\sum_{n=1}^\infty(-1)^n\frac{1}{n}$; taking either only the positive terms or the negative terms gives a sequence which is (almost) just a scalar multiple of the harmonic series, so diverges, but the individual terms may become arbitrarily small.
Now, lets say we want to use the same collection of terms to sum to a particular value $x$. Let's assume $x$ is positive, although it doesn't really matter. We can start adding up positive terms until we get to something larger than $x$ (the positive terms sum to $\infty$, so we won't run out before this happens). Now we can start adding the negative terms to move back towards $x$ again, until we overshoot slightly. Then add positive terms, etc. The conditions ensure (although it takes some care to see this formally) that
a) you never run out of terms of the right sign and get stuck on one side of $x$ or the other, and
b) that the terms you are using get smaller, so the error after each bit of adding becomes smaller in the long run.
