Easiest example of rearrangement of infinite leading to different sums I am reading the section on the rearrangement of infinite series in Ok, E. A. (2007). Real Analysis with Economic Applications. Princeton University Press.
As an example, the author shows that

is a rearrangement of the sequence
\begin{align} \frac{(-1)^{m+1}}{m} \end{align}
and that the infinite sum of these two sequences must be different.
My question is : what is to you the easiest and most intuitive example of such infinite series having different values for different arrangements of the terms? Ideally, I hope to find something as intuitive as the illustration that some infinite series do not have limits through $\sum_\infty (-1)^i$.
I found another example in http://www.math.ku.edu/~lerner/m500f09/Rearrangements.pdf but it is still too abstract to feed my intuition...

The following variant of the alternating harmonic is computationally easy. Consider the series
$$ 1-1+\frac{1}{2}-\frac{1}{2}+\frac{1}{2}-\frac{1}{2}+\frac{1}{4}-\frac{1}{4}+\frac{1}{4}-\frac{1}{4}+\frac{1}{4}-\frac{1}{4}+\frac{1}{4}-\frac{1}{4}+\frac{1}{8}-\frac{1}{8}+\cdots.$$
The sum is $0$. For the partial sums are either $0$ or $\frac{1}{2^k}$ for suitable $k$ that go to infinity.
Let us rearrange this series to give sum $1$.  Use 
$$\begin{align} 1+&\frac{1}{2}+\frac{1}{2}-1+\frac{1}{4}+\frac{1}{4}-\frac{1}{2}+\frac{1}{4}+\frac{1}{4}-\frac{1}{2}+\frac{1}{8}+\frac{1}{8}-\frac{1}{4}+\frac{1}{8}+\frac{1}{8}-\frac{1}{4}+\frac{1}{8}+\frac{1}{8}-\frac{1}{4}\\&+\frac{1}{8}+\frac{1}{8}-\frac{1}{4}+\frac{1}{16}+\frac{1}{16}-\frac{1}{8}+\frac{1}{16}+\frac{1}{16}-\frac{1}{8}+\cdots.\end{align}$$
 That the series converges to $1$ follows from the fact that the sum of the first $3n+1$ terms is always $1$, and that the sum of the first $3n+2$ terms, and the sum of the first $3n+3$ terms, differ from the sum of the first $3n+1$ terms by an amount that $\to 0$ as $n\to\infty$. 
