Maybe or maybe not. If you follow through the proof that rearranging the conditionally convergent series can make the sum anything you want, an important point is that the sum of the positive terms diverges to $+\infty$ and the sum of the negative terms diverges to $-\infty$. If that happens, you can use the same style of rearrangement. If not, consider the series $$a_n=\begin {cases} 2^{-n}&n \text { odd}\\ -1& n \text { even} \end {cases}$$ This is a diverging alternating series, but you cannot make it converge to anything. If you insist that the terms go to zero in absolute value (as required in the alternating series theorem) you can use $$a_n=\begin {cases} 2^{-n}&n \text { odd}\\ -1/n& n \text { even} \end {cases}$$
Consider the following
$$\sum_{n=1}^\infty \frac 1n =\infty,\qquad \sum_{n=1}^\infty \frac 1{2n}=\infty,\qquad \sum_{n=1}^\infty \frac 1{2n-1}=\infty,\qquad \sum_{n=1}^\infty \frac{(-1)^{n-1}}n=\ln(2).$$
In short: the sum of the reciprocals of all numbers diverges (harmonic series). The sum of the reciprocals of even/odd numbers diverges. But if you sum up even and odd reciprocals with alternating signs, it converges.
So here you have a specific infinite sum that seems to converge:
$$1-\frac12+\frac13-\frac14+\cdots =\ln(2).$$
Of course, if you just rearrange finitely many of the summands, you will end up with the same sum. But if you shuffle up all the numbers, you can get everywhere with your limit. I think this is done in any proof of Riemann's rearrangement theorem, but let me line out the proof on this example. Lets say you want the rearranged sum to converge to $\pi$ (for fun). Then take some of the positive terms (the odd reciprocals) and add enough of them up until you are just greater than $\pi$:
$$1+\frac13+\frac15+\cdots>\pi.$$
You can do this, as we know that the sum of the odd reciprocals diverges. In the next step, only take the negative (even) reciprocals and subtract them from your sum until you are just below $\pi$. Again you can do this, because the sum of the even reciprocals diverges. Now again take positiv terms, then negatives, then positives, and so on. In "the end" you will have used all the terms of the original sum, but you rearranged them in a way, so that they converge to $\pi$. And there is nothing special about $\pi$, so you can use this method to converge to anything, including $\pm\infty$.
Here a description of how to rearrange the sum to make it divergent, e.g. divergent to $\infty$. Sum up enough positive (odd) terms to make the sum greater than $1$. Add only a single negative term. Add positive terms until the sum exceed $2$. Add a single negative term. Add positive terms to exceed $3$, ... and so on. You will exceed any natural number, hence diverge to $\infty$.
Best Answer
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.