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Our professor gave a problem asking to rearrange the alternating harmonic series: $\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} $
such that the rearrangement equals infinity.

So I was doing some searching and found this property that the rearranged sums of the alternating harmonic series sum to:

$\ln(2) + \frac{1}{2}\ln(\frac{p}{n}) $

Where $p$ is the number of positive terms listed followed by $n$ negative terms in the rearrangement.

So my idea is that in order for the rearrangement to go to infinity, either $p$ is going to have to be infinite, or $n$ is going to have to be $0$. Would this make sense for the problem? It almost seems like this would not be a valid rearrangement of the alternating harmonic series, since I would basically be rearranging it to be the normal harmonic series.

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  • $\begingroup$ For those interested in historical origins of mathematical results, this particular result is due to Martin [Marcin, Martinus] Ohm (1792-1872) and I'm fairly sure it was first published in §8 (pp. 12-14) of Ohm's booklet De Nonnullis Seriebus Infinitis Summandis [Concerning the Summation of Certain Infinite Series], Trowitzschii et Filii [Trowitzsch und Sohn; Trowitzsch and Son] (Berlin), 1839, 15 pages. $\endgroup$ Feb 8, 2020 at 9:01

2 Answers 2

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Well. First, note that \begin{align} \sum^\infty_{n=1} \frac{1}{2n} =\infty. \end{align} Then we see that there exists $N_1$ such that \begin{align} 2<\frac{1}{2}+\frac{1}{4}+\ldots +\frac{1}{2N_1}<3 \end{align} then \begin{align} 1<\frac{1}{2}+\frac{1}{4}+\ldots +\frac{1}{2N_1}-1<2. \end{align} Next, we can find an $N_2$ such that \begin{align} 3<\frac{1}{2}+\frac{1}{4}+\ldots +\frac{1}{2N_1}-1+\frac{1}{2(N_1+1)}+\ldots+\frac{1}{2N_2}<4 \end{align} then \begin{align} 2<\frac{1}{2}+\frac{1}{4}+\ldots +\frac{1}{2N_1}-1+\frac{1}{2(N_1+1)}+\ldots+\frac{1}{2N_2}-\frac{1}{3}. \end{align} Again, choose $N_3$ such that \begin{align} 4<\frac{1}{2}+\frac{1}{4}+\ldots +\frac{1}{2N_1}-1+\frac{1}{2(N_1+1)}+\ldots+\frac{1}{2N_2}-\frac{1}{3}+\frac{1}{2(N_2+1)}+\ldots+\frac{1}{2N_3}<5 \end{align} which means \begin{align} 3<\frac{1}{2}+\frac{1}{4}+\ldots +\frac{1}{2N_1}-1+\frac{1}{2(N_1+1)}+\ldots+\frac{1}{2N_2}-\frac{1}{3}+\frac{1}{2(N_2+1)}+\ldots+\frac{1}{2N_3}-\frac{1}{5}. \end{align} Applying this process, you will obtain a rearrangment of $\sum (-1)^n/n$ such that the resulting series diverges.

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Define a rearrangement of the alternating harmonic series in blocks: Block 1 consists of the first $b_1$ positive (odd-denominator) terms of the AHS followed by the first negative (even-denominator) term. Block 2 has the next $b_2$ positive terms followed by the second negative term, and so on. For example, $b_1=1, b_2=3, b_3=2,\ldots$ corresponds to the arrangement: $$\frac11-\frac12+\frac13+\frac15+\frac17-\frac14+\frac19+\frac1{11}-\frac16+\cdots $$

Claim: If each $b_n\ge1$ and $(b_1+\cdots+b_n)/n\to\infty$, then the partial sums ($s_n$) of this arrangement diverge to infinity.

Proof: By construction, each block consists of one or more positive terms followed by one negative term. This implies that $s_n\ge s_{n-1}$ whenever $n$ does not correspond to the end of a block. Therefore it's enough to prove divergence along the 'blocked' subsequence.

Indeed, summing the first $n$ blocks of this rearranged series will consume the first $B_n:=b_1+\cdots+b_n$ positive terms and the first $n$ negative terms of the AHS: $$ s_{B_n+n} = \sum_{k=1}^{B_n}\frac1{2k-1}-\sum_{k=1}^n\frac1{2k}$$ The above expression is at least as big as: $$\sum_{k=1}^{B_n}\frac1{2k}-\sum_{k=1}^n\frac1{2k} =\frac12\sum_{n+1}^{B_n}\frac1k. $$ The conclusion follows from the inequality $$\sum_a^b\frac1k\ge\log\left(\frac{b+1}a\right).$$

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