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...

  • $\begingroup$ It may be easier for future questioners to see this if you mentioned conditionally converging series. $\endgroup$
    – Loki Clock
    Jun 6, 2013 at 19:04
  • $\begingroup$ 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. $\endgroup$ Jun 7, 2013 at 7:33
  • $\begingroup$ See Riemann series theorem. $\endgroup$
    – Loki Clock
    Jun 7, 2013 at 7:41

3 Answers 3


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

  • $\begingroup$ Nice! The rearrangement function even has an explicit formulation, the second sequence being $\frac{1}{\sigma(i)}$, where $\sigma(i) = \begin{cases} i - (2 - \frac{i}{2}), & \text{ when } i \text{ is even}\\ i - \frac{i-1}{2}, & \text{ when } i \text{ is odd and} \frac{i +(i-2)}{4} \text{ is odd} \\ i - \frac{i-3}{2}, & \text{ when } i \text{ is odd and} \frac{i +(i-2)}{4} \text{ is even} \\ \end{cases}$ $\endgroup$ Jun 6, 2013 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

  • $\begingroup$ While I am not sure about how intuitive it might seem, I find the argument solid and understandable $\endgroup$
    – Triatticus
    Jun 6, 2013 at 9:47
  • $\begingroup$ And now that I look at what you referenced I see its the same example...I suppose it's that popular $\endgroup$
    – Triatticus
    Jun 6, 2013 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 $n\to\infty$.

  • $\begingroup$ This argument is incomplete. The latter series has to converge to a different value, and hence has to be independent of groupings. $\endgroup$
    – Loki Clock
    Jun 6, 2013 at 17:27
  • $\begingroup$ 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. $\endgroup$ Jun 6, 2013 at 18:55

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