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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

enter image description here

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...

Martin Van der Linden
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3 Answers3

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

Gerry Myerson
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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 06 '13 at 11:29
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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

Triatticus
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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 $n\to\infty$.

André Nicolas
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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 06 '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 06 '13 at 18:55