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.
Best Answer
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.