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This problem comes from a Brazilian book of Real Analysis.

Explicitly perform a reordering of the terms of the series $$1 -\frac{1}{2} + \frac{1}{3} -\frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \frac{1}{7} - \frac{1}{8} + \cdots $$ so that its sum becomes zero.

It suggests to use the following reordering:

$$1 - \frac{1}{2} - \frac{1}{4} - \frac{1}{6} - \frac{1}{8} + \frac{1}{3} - \frac{1}{10} - \frac{1}{12} - \frac{1}{14} - \frac{1}{16} + \frac{1}{5} - \frac{1}{18} - \frac{1}{20} - \frac{1}{22}- \frac{1}{24} + \cdots$$

Where after each positive term is placed the first 4 negatives not used yet. But I don't know how to prove that this arrangement of the series converge to zero.

Rafael Deiga
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    Did it actually say using four? Without that restriction, it's easy enough - just precede each odd reciprocal by enough negative even reciprocals to make the bracketed sum negative. You can keep doing that, and can show that you can get to zero - the sum of the even reciprocals diverges, so you can define it in brackets so you add enough negative fractions together with the positive fraction to create a sum similar to 1 - (1/2) - (1/4) - (1/8) - (1/16) etc. Does that make sense? I haven't fully explained but can go more in depth if you don't understand – Franz Aug 19 '17 at 00:20
  • @Franz, sadly the answer is one which is easily understood by those who know it. I was taught the same one, but I think it could still be done more explicitly. An analogy of what Franz said is two have two bowls of numbers, (negative) even reciprocals and (positive) odd reciprocals, and then start with +1, "then add negative until you are negative, then add positive until you are positive" times infinity, continue until you are at 0. – mdave16 Aug 19 '17 at 00:25
  • @Franz I already know this result. I know that you can reorder the terms of a conditionally convergent series to get any real value that you want. But the question asks to you do it this explicitly. – Rafael Deiga Aug 19 '17 at 00:34

2 Answers2

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Yes, the suggested rearrangement converges to 0. This is a particular case of a result of Martin Ohm:

For $p$ and $q$ positive integers rearrange the sequence $$\left(\frac{(−1)^{n-1}} n\right)_{n\ge 1} $$ by taking the first $p$ positive terms, then the first $q$ negative terms, then the next $p$ positive terms, then the next $q$ negative terms, and so on. The rearranged series converges to $$ \ln(2) + \frac12 \ln\left( \frac pq \right). $$

There are more general explicit results along these lines, see for instance here. In the specific case that concerns us, $p=1$ and $q=4$, so the series converges to $\ln(2)+\frac12\ln(1/4)=\ln2-\ln2=0$.

The proof of Ohm's theorem is actually simple. You can find a quick presentation here, but I suggest trying on your own first. Here is a suggestion: For the case at hand, consider the partial sum of the rearrangement consisting of precisely $n$ blocks, where each block consists of adding the 1 odd term and subtracting 4 even ones, so the first block is $$ 1-\frac12-\frac14-\frac16-\frac18,$$ the next one is $$ 1-\frac12-\frac14-\frac16-\frac18+\frac13-\frac1{10}-\frac1{12}-\frac1{14}-\frac1{16},$$ etc. Easy algebra shows this equals $$S_{8n}-\sum_{k=n}^{4n-1}\frac1{2k+1},$$ where $S_l$ is the $l$th partial sum of the alternating harmonic series. We have $S_{8n}\to\ln2$ (see here). Evaluate the second expression by comparing it to $$\frac12\int_{n}^{4n}\frac1x\,dx.$$

By the way, this is the rearrangement obtained by following the usual proof of Riemann's theorem. One can check that following Riemann's recipe for the alternating harmonic series to obtain a rearrangement that converges to 0, indeed one always adds exactly one positive (odd) term, then subtracts several even terms, adds the next odd term, subtracts several even ones, etc, but finding the precise pattern is a bit more involved. For details, see

MR2663251. Freniche, Francisco J. On Riemann's rearrangement theorem for the alternating harmonic series. Amer. Math. Monthly 117 (2010), no. 5, 442–448.

(In particular, theorem 3 and remark 4 at the end, that indicate the eventual pattern is indeed 1 positive term, then 4 negative ones.)

Andrés E. Caicedo
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Its not actually four terms. Perhaps, to make the hint more obvious:

$$\begin{align}0&<1\\0&>1-\frac12-\frac14-\frac16-\frac18\\0&<1-\frac12-\frac14-\frac16-\frac18+\frac13\\0&>1-\frac12-\frac14-\frac16-\frac18+\frac13-\frac1{10}-\frac1{12}-\frac1{14}-\frac1{16}\end{align}$$

If you rearrange it so that the partial sums always go above zero and then below zero, and forever alternating in this manner, what should the limit be?

And can you prove that you can always pull the partial sums to the other side of zero? Hint: The harmonic series diverges.

Simply Beautiful Art
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