Rearrangement in series I no understanding of rearrangement.
Why are we doing rearrangement?
Does it mean some divergent series have convergent series in rearrangement? 
How does this make sense?
To me 1, 2, 3, 4 is exactly the same as 3,4,1,2
I probably not understood any from rearrangement. Please give some basic idea of this to begin with. 
 A: If you have a finite sum, then rearrangements are trivial. It's when you get into the infinite that you begin to see some strange things happen.
For absolutely convergent series, we get back a series that, again, acts the way we would like. It's only when we go with a rather loose concept of convergence, that is, conditional convergence, that we see strange things happen.
I think the typical example is the alternating harmonic series, $\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n+1}}{n}$. In this original arrangement, the sum is $\ln2$. However, we may rearrange it for another sum as follows:
$$\left(1-\frac{1}{2}\right)-\frac{1}{4}+\left(\frac{1}{3}-\frac{1}{6}\right)-\frac{1}{8}+\left(\frac{1}{5}-\frac{1}{10}\right)-\frac{1}{12}+\cdots=$$
$$\frac{1}{2}-\frac{1}{4}+\frac{1}{6}-\frac{1}{8}+\frac{1}{10}-\frac{1}{12}+\cdots=$$
$$\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n+1}}{2n}=\frac{1}{2}\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n+1}}{n}=\frac{1}{2}\ln2.$$
What we did was move a few terms over and see that we somehow got half of our original sum. This is the Riemann Rearrangement Theorem. In fact, any conditionally convergent series can be made to become anything, essentially. We can obtain any real number or even make a divergent rearrangement.
Here's a nonrigorous reason for why this is true. Note that a conditionally convergent series is that way because the absolute value of the $n$th term isn't going to 0 fast enough, and so both the positive and negative terms in such a series, on their own, would be infinite (so we can add only positive or only negative to pass any value in a finite number of terms). So, when we rearrange to get some real number $x$, which we will assume is positive without loss in generality (if negative, swap the words positive/above and negative/below in the next bit), we can add the positive terms up until we get to or above $x$, and then we add negative terms until we get below, and continue in that fashion. Since the $n$th term is going to zero eventually, our difference from $x$ gets smaller the further we take this process so that, in the limit, we get $x$.
If we want a divergent sum, we add a bunch of positive values up and then a few negative values, then go back to positive, and return to negative for a short time. I'm being very vague here, but since the positive and negative terms, on their own, diverge (conditional convergence), we can add the terms in such a way that the positive terms are dominating the few negative terms that we put in at a time so that the limit is infinite. We can, of course, do the same the other way around.
The reason that this doesn't work with absolute convergent series is due to the fact that the $n$th term is going to 0 fast enough. We could never add the positive terms to get an arbitrarily large number, and the negative terms that we do add, no matter how cleverly we may think that we're adding them, will be enough to put our sum back in it's place. The $n$th term of an absolutely convergent series is simply too weak to move the series away from it's original sum.
A: Note two things in rearrangement of series:


*

*We must have "many", i.e. infinitely, negative terms. Same goes for positive terms.

*The original series is convergent. 


A consequence of these is that we can choose as many small terms as we wish and still have extra in reserve.

Since you took a finite example and got confused by that, consider the following (not mathematically precise) one.
Suppose that we are running some business and revenues are given in terms of
$$1,-1,1,-1,\dots,-1,1,$$
where we have, say a million of $-1$'s and a million and one of $1$'s. Of course, at the end, our business is profitable. However, rearranging revenues makes a huge difference. If it happens that the first few hundred thousand terms were $-1$, our business is really bad and we may get bankruptcy before making a profit. 
Similarly, by choosing $1$ or $-1$, one can rearrange the above sequence to make any (whole) number if we just look at the first half of the sum. 
The idea of the proof (See Brian's answer or maybe your textbook) is pretty much the same, except that:


*

*We now deal with infinitely many terms (the sum never stops)

*We have many small terms (much smaller than 1) to add/subtract at each step.

*We may not get precisely the wanted number at each step, but that where limit comes in.

