Rearrangement of Alternating Harmonic Series to be Infinity

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.

• 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. Feb 8, 2020 at 9:01

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.
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 $$\sum_a^b\frac1k\ge\log\left(\frac{b+1}a\right).$$