In my first course in analysis, we briefly covered the proof of the Riemann Rearrangement theorem as an exercise, from which we are assured that we could rearrange the alternating harmonic series to converge to, for example, $\pi$.
However, the rearrangement required to achieve $\pi$ in this way would be quite contrived - I'm assuming that there would not be any obvious pattern to this rearrangement. I'm curious to know whether or not there are many known "natural" rearrangements with more sensible patterns.
The linked Wikipedia article gives one quite nice natural rearrangement:
Suppose that two positive integers $a$ and $b$ are given, and that a rearrangement of the alternating harmonic series is formed by taking, in order, $a$ positive terms from the alternating harmonic series, followed by $b$ negative terms, and repeating this pattern at infinity [...]
It follows that the sum of this rearranged series is: $$\ln\bigg({2\sqrt\frac{a}{b}}\bigg)$$
Are there any other such natural rearrangements (for any conditionally convergent series) that give nice results?
Let me get you started, since I judge that your comment means you don't know how to begin. We rearrange the series $$1-\frac13+\frac15-\frac17+\cdots$$ by adding two positive terms and then one negative term, ad infinitum. $$S=1+\frac15-\frac13+\frac19+\frac1{13}-\frac17$$ Separate the sum into groups of three: $$S=\left(1+\frac15-\frac13\right)+\left(\frac19+\frac1{13}-\frac17\right)+\cdots+\left(\frac1{8n+1}+\frac1{8n+5}-\frac1{4n+3}\right)+\cdots$$
A moment's thought shows that if this series converges, so does the original series, to the same sum. The partial sums are the subsequence $s_{3n}$ of the sequence $s_n$ of partial sums of the original series. Since the terms go to $0$ if $s_{3n}\to S$ then also $s_{3n+1}\to S$ and $s_{3n+2}\to S$ and then $s_n\to S$.
We have $$\begin{align}s_{3n} &=\sum_{k=0}^n\left(\frac1{8k+1}+\frac1{8k+5}\right)-\sum_{k=0}^n\frac1{4k+3}\\ &=\sum_{k=0}^{2n+1}\frac1{4k+1}- \sum_{k=0}^n\frac1{4k+3}\\ &=\sum_{k=0}^n\left(\frac1{4k+1}-\frac1{4k+3}\right) +\sum_{k=n+1}^{2n+1}\frac1{4k+1} \end{align}$$
As $n\to\infty$, we know that the first sum $\to\frac\pi4$, so that we need to evaluate $$t:=\lim_{n\to\infty}\sum_{k=n+1}^{2n+1}\frac1{4k+1}$$ It's easy to see that the limit exists and that $\frac18\leq t\leq\frac14$.
This is as far as I've gone, and I'm going to turn it over to you now, with the suggestion that you might try to relate this to the harmonic numbers. (I'm going to guess $t=\frac\gamma4$, where $\gamma$ is Euler's constant, $\gamma\approx.5772$, but that may be wildly wrong.)
I think if you can do this case, you'll be able to generalize it to other rearrangements of the same sort.