Riemann's Rearrangement Theorem

Termwise $a_n = p_n - q_n$ and $\lvert a_n \rvert = p_n + q_n$ (zero terms contribute nothing, so this is the positive/negative subsequences re-indexed). Conditional convergence means $\sum (p_n - q_n)$ converges while $\sum (p_n + q_n)$ diverges.

Suppose $\sum p_n < \infty$. Since $\sum a_n = \sum (p_n - q_n)$ converges, and $q_n = p_n - (p_n - q_n)$, the series $\sum q_n = \sum p_n - \sum a_n$ converges as a difference of convergent series.

Then $\sum \lvert a_n \rvert = \sum p_n + \sum q_n < \infty$ — so $\sum a_n$ would converge absolutely, contradicting conditional convergence. Hence $\sum p_n = +\infty$.

Symmetrically: if $\sum q_n < \infty$, then $p_n = q_n + (p_n - q_n)$ gives $\sum p_n = \sum q_n + \sum a_n$ convergent, again forcing $\sum \lvert a_n \rvert < \infty$ — contradiction.

So $\sum q_n = +\infty$.

The intuition: a conditionally convergent series harbors both "positive infinity" and "negative infinity", and a rearrangement can shuttle between them to hit any target.

The rule: take the next positive term while $t \le s$, the next negative term while $t > s$. For $s = 1$ with $t_0 = 0$, the first take is positive: $S_1 = 0 + p_1 = 1$ — the walk lands exactly on the target (gap 0).

Since $S_1 = 1 \le 1$, one more positive: $S_2 = 1 + p_2 = 1 + \frac13 = \frac43$ — the first positive block ends with overshoot $\frac13 = p_2$ (equality holds when crossing upward from exactly $s$; see the lemma below). Then $S_3 = \frac43 - q_1 = \frac43 - \frac12 = \frac56$ — the first negative block, undershoot $\frac16 < q_1 = \frac12$.

Continuing: $S_4 = \frac56 + p_3 = \frac56 + \frac15 = \frac{31}{30}$ (overshoot $\frac1{30} < p_3 = \frac15$), $S_5 = \frac{31}{30} - q_2 = \frac{31}{30} - \frac14 = \frac{47}{60}$ (undershoot $\frac{13}{60} < q_2 = \frac14$).

The next positive block takes two terms: $S_6 = \frac{47}{60} + p_4 = \frac{389}{420} < 1$ (not yet across), $S_7 = \frac{389}{420} + p_5 = \frac{1307}{1260}$ (overshoot $\frac{47}{1260} < p_5 = \frac19$).

And $S_8 = \frac{1307}{1260} - q_3 = \frac{1097}{1260}$ (undershoot $\frac{163}{1260} < q_3 = \frac16$).

Every value matches the model computation (verified as exact rationals in the tests).

The gap sequence $\frac13, \frac16, \frac1{30}, \frac{13}{60}, \frac{47}{1260}, \frac{163}{1260}$ is not monotone, but each gap is at most its crossing term, and the crossing terms tend to 0.

After 120 terms the walk sits at $S_{120} \approx 0.9967$ with error $\approx 0.00326$ — the bracket is closing in on the target.

The alternating harmonic series $\sum \frac{(-1)^{n+1}}{n}$ converges conditionally (Leibniz gives convergence; $\sum \frac1n$ diverges, so no absolute convergence). For the positive terms $p_k = \frac{1}{2k-1}$ and negative magnitudes $q_k = \frac{1}{2k}$, we have $\sum p_k = \sum q_k = +\infty$ (a corollary of the previous problem, or directly via $\frac{1}{2k-1} \ge \frac{1}{2k}$ and comparison with the harmonic series).

Both blocks recur forever.

If positives were taken only finitely often, then eventually only negatives are taken: the running sum drops by $q_j$ each step and $\sum q_j = \infty$ forces $t \to -\infty$ — eventually $t \le s$, and the rule would take a positive: contradiction.

Symmetrically for negatives (otherwise $t \to +\infty$ would force a negative take).

Since each block consumes the next unused terms in order, every $p_i$ and every $q_j$ is eventually used exactly once — a genuine rearrangement.

Crossing lemma: crossing upward from $t \le s$ by adding $p$, the new gap is $(t + p) - s = p - (s - t) \le p$ (equality only when $t = s$).

Crossing downward from $t > s$ by subtracting $q$, the new gap is $s - (t - q) = q - (t - s) < q$.

So every crossing overshoot/undershoot is at most its crossing term's magnitude.

Now look at one positive block (after the first crossing; finitely many initial steps don't affect convergence).

Starting from an undershoot gap $\varepsilon$ (with $\varepsilon \le q_{\text{last}}$, the last negative taken, by the lemma), the block increases monotonically, so before the crossing step $s - S \le \varepsilon$, and at the crossing step $S - s \le p_{\text{cross}}$.

Hence throughout the block $\lvert S_N - s \rvert \le \max(q_{\text{last}}, p_{\text{cross}})$.

Since the blocks recur forever (step 2), both term indices tend to $\infty$: $q_{\text{last}} \to 0$ and $p_{\text{cross}} \to 0$ (the crossing positives are a subsequence of the $p_i$).

Hence $\lvert S_N - s \rvert \to 0$, i.e. $S_N \to s$.

Negative blocks are symmetric.

This is the greedy proof of Riemann's rearrangement theorem for finite $s$.

(The $\pm\infty$ cases lie outside this statement's scope and need a separate construction: for target $+\infty$, for instance, replace the fixed target $s$ by stage targets $m \to +\infty$ — at stage $m$, take positive terms until the partial sum exceeds $m$.)