# 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...

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It may be easier for future questioners to see this if you mentioned conditionally converging series. – Loki Clock Jun 6 '13 at 19:04
I am not familiar with the notion of conditionally converging series. Is it that I should replace any instance of "infinite series" by "conditionally converging series"? Anyways, fell free to edit my question if you have an improvement. – Martin Van der Linden Jun 7 '13 at 7:33
– Loki Clock Jun 7 '13 at 7:41

$$1/2-1/3+1/4-1/5+1/6-1/7+\cdots=(1/2-1/3)+(1/4-1/5)+(1/6-1/7)+\cdots$$ 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 $$(1/2-8/15)+(1/4-16/63)+(1/6-24/143)+\cdots\lt(1/2-8/16)+(1/4-16/64)+(1/6-24/144)+\cdots=0$$

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Nice! The rearrangement function even has an explicit formulation, the second sequence being $\frac{1}{\sigma(i)}$, where $\sigma(i) = \begin{cases} i - (2 - \frac{i}{2}), & \text{ when } i \text{ is even}\\ i - \frac{i-1}{2}, & \text{ when } i \text{ is odd and} \frac{i +(i-2)}{4} \text{ is odd} \\ i - \frac{i-3}{2}, & \text{ when } i \text{ is odd and} \frac{i +(i-2)}{4} \text{ is even} \\ \end{cases}$ – Martin Van der Linden Jun 6 '13 at 11:29

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

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While I am not sure about how intuitive it might seem, I find the argument solid and understandable – Triatticus Jun 6 '13 at 9:47
And now that I look at what you referenced I see its the same example...I suppose it's that popular – Triatticus Jun 6 '13 at 9:48

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 $n\to\infty$.

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This argument is incomplete. The latter series has to converge to a different value, and hence has to be independent of groupings. – Loki Clock Jun 6 '13 at 17:27
There is no grouping, the "triplets" remark is just a description of how the sequence is built. The convergence to $1$ is obvious, since the sum of the first $3k+1$ elements is always $1$, and for large $k$ the $3k+2$-th sum and the $3k+3$-th sum differ from $1$ by an amount that $\to 0$. But I will change the sentence at the end since it may cause confusion. – André Nicolas Jun 6 '13 at 18:55