Riemann's Rearrangement Theorem

A conditionally convergent series can be rearranged to sum to any real number — watch the greedy block rule steer the alternating harmonic series (which sums to ln⁡2\ln 2 in its natural order) onto a target of your choice.

∑an converges conditionally ⟹ ∀ s∈R  ∃ rearrangement with sum s\sum a_n\ \text{converges conditionally}\ \Longrightarrow\ \forall\, s\in\mathbb{R}\ \ \exists\ \text{rearrangement with sum } s

Part of these learning paths

Examples & Exercises

Let ∑an\sum a_n be a real series converging conditionally. With pn=max⁡(an,0)p_n = \max(a_n, 0) and qn=max⁡(−an,0)q_n = \max(-a_n, 0) (zero terms are harmless, so this is the positive/negative subsequences re-indexed), prove ∑pn=+∞\sum p_n = +\infty and ∑qn=+∞\sum q_n = +\infty.

an=pn−qn,∣an∣=pn+qna_n = p_n - q_n, \qquad \lvert a_n \rvert = p_n + q_n
∑pn<∞ ⟹ ∑qn=∑pn−∑an<∞ ⟹ ∑∣an∣<∞(contradiction)\sum p_n < \infty \ \Longrightarrow\ \sum q_n = \sum p_n - \sum a_n < \infty \ \Longrightarrow\ \sum \lvert a_n \rvert < \infty \quad \text{(contradiction)}
Target s=1s = 1, N=120N = 120: S120≈0.9967S_{120} \approx 0.9967, error ≈0.00326\approx 0.00326 — 78 positive and 42 negative takes, and both blocks recur forever (the proof below is why).
Show proof
  1. Termwise an=pn−qna_n = p_n - q_n and ∣an∣=pn+qn\lvert a_n \rvert = p_n + q_n (zero terms contribute nothing, so this is the positive/negative subsequences re-indexed). Conditional convergence means ∑(pn−qn)\sum (p_n - q_n) converges while ∑(pn+qn)\sum (p_n + q_n) diverges.
  2. Suppose ∑pn<∞\sum p_n < \infty. Since ∑an=∑(pn−qn)\sum a_n = \sum (p_n - q_n) converges, and qn=pn−(pn−qn)q_n = p_n - (p_n - q_n), the series ∑qn=∑pn−∑an\sum q_n = \sum p_n - \sum a_n converges as a difference of convergent series.
  3. Then ∑∣an∣=∑pn+∑qn<∞\sum \lvert a_n \rvert = \sum p_n + \sum q_n < \infty — so ∑an\sum a_n would converge absolutely, contradicting conditional convergence. Hence ∑pn=+∞\sum p_n = +\infty.
  4. Symmetrically: if ∑qn<∞\sum q_n < \infty, then pn=qn+(pn−qn)p_n = q_n + (p_n - q_n) gives ∑pn=∑qn+∑an\sum p_n = \sum q_n + \sum a_n convergent, again forcing ∑∣an∣<∞\sum \lvert a_n \rvert < \infty — contradiction. So ∑qn=+∞\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.

Aim the scene's greedy walk at target s=1s = 1. Compute the partial sums S2,S3,S4,S5,S7,S8S_2, S_3, S_4, S_5, S_7, S_8 at the ends of the first few blocks by hand, and check that each overshoot/undershoot is at most the magnitude of the term that caused the crossing.

S2=43, S3=56, S4=3130, S5=4760, S7=13071260, S8=10971260S_2 = \tfrac43,\ S_3 = \tfrac56,\ S_4 = \tfrac{31}{30},\ S_5 = \tfrac{47}{60},\ S_7 = \tfrac{1307}{1260},\ S_8 = \tfrac{1097}{1260}
gaps: 13, 16, 130, 1360, 471260, 1631260 ≤ crossing term each time\text{gaps: }\tfrac13,\ \tfrac16,\ \tfrac1{30},\ \tfrac{13}{60},\ \tfrac{47}{1260},\ \tfrac{163}{1260}\ \le\ \text{crossing term each time}
Target s=1s = 1, N=120N = 120: the bracket tightens 13→16→130→1360→471260→1631260\frac13 \to \frac16 \to \frac1{30} \to \frac{13}{60} \to \frac{47}{1260} \to \frac{163}{1260}, reaching S120≈0.9967S_{120} \approx 0.9967 with error ≈0.00326\approx 0.00326.
Show proof
  1. The rule: take the next positive term while t≤st \le s, the next negative term while t>st > s. For s=1s = 1 with t0=0t_0 = 0, the first take is positive: S1=0+p1=1S_1 = 0 + p_1 = 1 — the walk lands exactly on the target (gap 0).
  2. Since S1=1≤1S_1 = 1 \le 1, one more positive: S2=1+p2=1+13=43S_2 = 1 + p_2 = 1 + \frac13 = \frac43 — the first positive block ends with overshoot 13=p2\frac13 = p_2 (equality holds when crossing upward from exactly ss; see the lemma below). Then S3=43−q1=43−12=56S_3 = \frac43 - q_1 = \frac43 - \frac12 = \frac56 — the first negative block, undershoot 16<q1=12\frac16 < q_1 = \frac12.
  3. Continuing: S4=56+p3=56+15=3130S_4 = \frac56 + p_3 = \frac56 + \frac15 = \frac{31}{30} (overshoot 130<p3=15\frac1{30} < p_3 = \frac15), S5=3130−q2=3130−14=4760S_5 = \frac{31}{30} - q_2 = \frac{31}{30} - \frac14 = \frac{47}{60} (undershoot 1360<q2=14\frac{13}{60} < q_2 = \frac14). The next positive block takes two terms: S6=4760+p4=389420<1S_6 = \frac{47}{60} + p_4 = \frac{389}{420} < 1 (not yet across), S7=389420+p5=13071260S_7 = \frac{389}{420} + p_5 = \frac{1307}{1260} (overshoot 471260<p5=19\frac{47}{1260} < p_5 = \frac19). And S8=13071260−q3=10971260S_8 = \frac{1307}{1260} - q_3 = \frac{1097}{1260} (undershoot 1631260<q3=16\frac{163}{1260} < q_3 = \frac16).
  4. Every value matches the model computation (verified as exact rationals in the tests). The gap sequence 13,16,130,1360,471260,1631260\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 S120≈0.9967S_{120} \approx 0.9967 with error ≈0.00326\approx 0.00326 — the bracket is closing in on the target.

Prove the greedy rearrangement converges to the target ss: both blocks recur forever, each crossing overshoot/undershoot is at most the crossing term's magnitude, and since the term magnitudes tend to 0, ∣SN−s∣→0\lvert S_N - s \rvert \to 0.

t≤s⇒S−s≤p(crossing up),t>s⇒s−S<q(crossing down)t \le s \Rightarrow S - s \le p \quad (\text{crossing up}), \qquad t > s \Rightarrow s - S < q \quad (\text{crossing down})
∣SN−s∣≤max⁡(qlast,pcross)  →N→∞  0\lvert S_N - s \rvert \le \max(q_{\text{last}}, p_{\text{cross}}) \;\xrightarrow[N\to\infty]{}\; 0
Target s=1s = 1, N=120N = 120: S120≈0.9967S_{120} \approx 0.9967, error ≈0.00326\approx 0.00326 — as the term magnitudes die out, so does the walk's amplitude around the target.
Show proof
  1. The alternating harmonic series ∑(−1)n+1n\sum \frac{(-1)^{n+1}}{n} converges conditionally (Leibniz gives convergence; ∑1n\sum \frac1n diverges, so no absolute convergence). For the positive terms pk=12k−1p_k = \frac{1}{2k-1} and negative magnitudes qk=12kq_k = \frac{1}{2k}, we have ∑pk=∑qk=+∞\sum p_k = \sum q_k = +\infty (a corollary of the previous problem, or directly via 12k−1≥12k\frac{1}{2k-1} \ge \frac{1}{2k} and comparison with the harmonic series).
  2. Both blocks recur forever. If positives were taken only finitely often, then eventually only negatives are taken: the running sum drops by qjq_j each step and ∑qj=∞\sum q_j = \infty forces t→−∞t \to -\infty — eventually t≤st \le s, and the rule would take a positive: contradiction. Symmetrically for negatives (otherwise t→+∞t \to +\infty would force a negative take). Since each block consumes the next unused terms in order, every pip_i and every qjq_j is eventually used exactly once — a genuine rearrangement.
  3. Crossing lemma: crossing upward from t≤st \le s by adding pp, the new gap is (t+p)−s=p−(s−t)≤p(t + p) - s = p - (s - t) \le p (equality only when t=st = s). Crossing downward from t>st > s by subtracting qq, the new gap is s−(t−q)=q−(t−s)<qs - (t - q) = q - (t - s) < q. So every crossing overshoot/undershoot is at most its crossing term's magnitude.
  4. Now look at one positive block (after the first crossing; finitely many initial steps don't affect convergence). Starting from an undershoot gap ε\varepsilon (with ε≤qlast\varepsilon \le q_{\text{last}}, the last negative taken, by the lemma), the block increases monotonically, so before the crossing step s−S≤εs - S \le \varepsilon, and at the crossing step S−s≤pcrossS - s \le p_{\text{cross}}. Hence throughout the block ∣SN−s∣≤max⁡(qlast,pcross)\lvert S_N - s \rvert \le \max(q_{\text{last}}, p_{\text{cross}}).
  5. Since the blocks recur forever (step 2), both term indices tend to ∞\infty: qlast→0q_{\text{last}} \to 0 and pcross→0p_{\text{cross}} \to 0 (the crossing positives are a subsequence of the pip_i). Hence ∣SN−s∣→0\lvert S_N - s \rvert \to 0, i.e. SN→sS_N \to s. Negative blocks are symmetric. This is the greedy proof of Riemann's rearrangement theorem for finite ss (the ±∞\pm\infty cases use the same machine with an ever-growing target).

7. Power Series