Riemann Rearrangement

In the usual order, 1 − ½ + ⅓ − ¼ + … adds up to ln 2. Choose a target, or a pattern of positive and negative terms, and watch the same terms walk to a different sum, one term at a time.

Worked example — Riemann Rearrangement

Positive terms used, P	99,259
Negative terms used, Q	741
Total after 100,000 terms	3.1415535
Positive terms per negative one	133.95
The ratio the target asks for	133.87
Usual order, after 100,000 terms	0.6931422

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Lesson

The theory — Riemann Rearrangement

A series converges conditionally when it converges but the sizes of its terms add up to infinity. Riemann showed that the terms of such a series can be re-ordered to add up to any number at all, while a series that still converges with every sign made positive has the same sum in every order.

What each symbol means

P, Q: how many positive and how many negative terms have been used, each half taken in its own order
C: the target Riemann’s recipe aims at
r: the long-run ratio P/Q; the total heads for ln 2 + ½ ln r

Where the formula comes from

Why the recipe arrives. Take positive terms while the total is at or below C, negative terms while it is above. Each half of 1 − ½ + ⅓ − … adds up to more than any number on its own, so however far the total has gone it always crosses C again. Just after a crossing it is no further from C than the term that carried it across, and the terms that follow only move it back towards C until the next crossing. The terms shrink to 0, so the distance does too.
Why the squares cannot be steered. When the sizes of the terms add up to a finite number, then for any ε, however small, the sizes beyond some point M add up to less than ε. A re-ordering that has used all of the first M terms differs from the usual M-th partial sum only by terms beyond M, so by less than ε, and so both orders have the same limit. The recipe fails from the other side: it needs an unlimited supply of terms of each sign, and here each supply is finite.

Re-order 1 − ½ + ⅓ − … and it adds up to π 🖖

Press Aim at π. The page takes positive terms 1, ⅓, ⅕, … while the running total is at or below π, and negative terms ½, ¼, ⅙, … while it is above, each half in its own order, so no term is used twice. After 100,000 terms the total is 3.1415535, from 99,259 positive terms against 741 negative ones. The grey line is the usual order, still at 0.6931422 and heading for ln 2. The recipe is Riemann’s, from the essay on trigonometric series he submitted at Göttingen in 1854 to qualify as a lecturer. The positive terms on their own add up to more than any number you can name, and so do the negative ones, so the total always crosses the target again. Once across, it is never further from the target than the size of the term that carried it over, and the terms shrink to nothing.

The sum is set by one ratio: positive terms spent per negative one 🖖

Take each half in its own order, spend positive and negative terms in the long-run ratio r, and the total heads for ln 2 + ½ ln r. Press One up, two down: the same terms as before, and the total closes on 0.3465736, exactly half of ln 2. Nine up and four down gives ln 2 + ½ ln(9/4), which is ln 3. Read backwards, the formula prices a target. π costs e^{2(π − ln 2)} = 133.87 positive terms per negative one, and the run’s 99,259 to 741 is 133.95. Press Aim at 0: the recipe knows nothing about patterns, and for all 100,000 terms it takes one positive and then four negatives, term for term: ln 2 + ½ ln ¼ = 0. The lower panel shows the share of positive terms settling where the target puts it.

Square the terms and no order can move the sum 🖖

Switch to 1 − ¼ + 1/9 − 1/16 + …. Its positive terms add up to only π²/8 = 1.2337006 and its negative ones to π²/24, so every order that uses all the terms ends at π²/12 = 0.8224670. Try any pattern. Riemann’s recipe now jams. Press Squares, aim at 1.5: all 100,000 terms are positive and the total stalls at 1.2336981. Aim at 0.8 instead and 1,000 terms look like success, 0.8000080; by 100,000 the total is stuck at 0.8016482, because the negative terms still unused add up to less than the 0.0016482 it is over. Riemann drew the dividing line exactly here and credited it to Dirichlet: a series that still converges when every sign is made positive can have its terms moved anywhere, and only the other kind depends on the order.

Problem solved in full

Which pattern adds up to ln 3, and which to 1.5? 4 steps
Re-order 1 − ½ + ⅓ − … into a repeating pattern of p positive terms and q negative ones that adds up to ln 3. Then find one that adds up to 1.5.

Open this problem in the tool
1. After P positive terms and Q negative ones, each half taken in order, the total is a sum of odd reciprocals minus half a harmonic sum, and both are harmonic sums H_n = 1 + ½ + … + 1/n in disguise:
  
  $S = (1 + \frac{1}{3} + \dots + \frac{1}{2 P - 1}) - \frac{1}{2} (1 + \frac{1}{2} + \dots + \frac{1}{Q}) = H_{2 P} - \frac{1}{2} H_{P} - \frac{1}{2} H_{Q}$
2. H_n is ln n + γ plus a remainder that shrinks to 0, and the three γ cancel:
  
  $H_{n} = ln n + γ + o (1) \Rightarrow S = ln 2 P - \frac{1}{2} ln P - \frac{1}{2} ln Q + o (1) \to ln 2 + \frac{1}{2} ln \frac{P}{Q}$
3. For ln 3 the ratio r = p/q must satisfy ½ ln r = ln 3 − ln 2:
  
  $ln 2 + \frac{1}{2} ln r = ln 3 \Rightarrow r = (\frac{3}{2})^{2} = \frac{9}{4} \Rightarrow ln 2 + \frac{1}{2} ln \frac{9}{4} = 1.0986123$
4. For 1.5 the same equation asks for a ratio no pattern can supply:
  
  $ln 2 + \frac{1}{2} ln r = 1.5 \Rightarrow r = \frac{e ^{3}}{4} = 5.0214 (irrational), 5 : 1 \to 1.4978661$
Answer

Nine positive terms, then four negative, add up to ln 3: the page prints 1.0986331 after 100,000 terms, closing on 1.0986123. No repeating pattern adds up to 1.5. The ratio it needs is e³/4 = 5.0214, and e³ is irrational because e is transcendental, so every whole-number pattern misses; 5 : 1 heads for 1.4978661. Riemann’s recipe gets there anyway, because its ratio is free to drift: aimed at 1.5 it spends 83,393 positive terms and 16,607 negative ones, 5.0216 per negative, and reaches 1.5000022.

$9 : 4 \to ln 3; 1.5 : no pattern, but greedy \frac{83 , 393}{16 , 607} = 5.0216 \to 1.5000022$

Learning path

Infinite series and convergence

References (2)

Riemann's own statement of the recipe the page runs, take positive terms until the total exceeds C and negative ones until it falls below, with the bound that the total never strays further than the term before the last change of sign; and the division, which he credits to Dirichlet, between series whose terms can be moved freely and series whose value depends on the order: B. Riemann, "Ueber die Darstellbarkeit einer Function durch eine trigonometrische Reihe", Habilitationsschrift, Göttingen 1854; published by R. Dedekind in Abhandlungen der Königlichen Gesellschaft der Wissenschaften zu Göttingen 13 (1867), §3. Transcribed by D. R. Wilkins, Trinity College Dublin.
The rearrangement one positive term then two negative ones, (1 − ½ − ¼) + (⅓ − ⅙ − ⅛) + …, summed to ½ ln 2: E. W. Weisstein, "Riemann Series Theorem", MathWorld, Wolfram Research.

By the 42Math team · Reviewed against published sources. Corrections welcome.