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...
 A: There is an argument in the Stewart calculus book 5th edition (the one I learned from) which uses the alternating harmonic series since it is conditionally convergent which goes like this:
The series:
$$ 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \dots =  \ln 2 $$
Multiplying the series by half yields:
$$ \frac{1}{2} - \frac{1}{4} + \frac{1}{6} - \frac{1}{8} +  \dots = \frac{1}{2} \ln 2 $$
He then does a trick by inserting zeros between each number:
$$0 + \frac{1}{2} +0 - \frac{1}{4}+0 + \frac{1}{6}+0 - \frac{1}{8}+ \dots = \frac{1}{2} \ln 2 $$
He then adds the original sum and the newly acquired sum above and obtains:
$$ 1 + \frac{1}{3} - \frac{1}{2} + \frac{1}{5} + \frac{1}{7} - \frac{1}{4} + \dots = \frac{3}{2} \ln 2 $$
He asserts that this is the original series with it's terms rearranged with pairwise positive terms followed by negatives yielding a completely different sum.
Stewart, J "Single-Variable Calculus" 5th edition
A: 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 $n\to\infty$. 
A: $$1/2-1/3+1/4-1/5+1/6-1/7+\cdots=(1/2-1/3)+(1/4-1/5)+(1/6-1/7)+\cdots$$ is obviously positive. The rearrangement $$1/2-1/3-1/5+1/4-1/7-1/9+1/6-1/11-1/13+\cdots$$ is clearly negative; just group it as $$(1/2-1/3-1/5)+(1/4-1/7-1/9)+(1/6-1/11-1/13)+\cdots$$ which is $$(1/2-8/15)+(1/4-16/63)+(1/6-24/143)+\cdots\lt(1/2-8/16)+(1/4-16/64)+(1/6-24/144)+\cdots=0$$
