Adding infinitely many numbers. Geometric series, a tower of blocks that leans arbitrarily far, convergence tests, and a series you can rearrange to any sum
What does it mean to add infinitely many numbers? Add the first N, see where the partial sums are heading as N grows, and call that limit the sum, if there is one. Sometimes the answer is a tidy number, sometimes the sums grow without end, and sometimes the answer depends on the order in which you add.
Fill a square with ever smaller halves, then slide the ratio r and see when the partial sums settle and when they run away. Zeno’s runner finishes infinitely many stages in one second.
In the ratio picture the vertical range is capped at ±60, so a divergent series runs off the top. The runner covers one track length per second, so stage k (from 1 − 2¹⁻ᵏ to 1 − 2⁻ᵏ) takes 2⁻ᵏ seconds.
A geometric series multiplies by the same ratio r at each step: a + ar + ar² + …. Its partial sums have a closed form. Write Sₙ = a + ar + … + arⁿ⁻¹, multiply by r and subtract: Sₙ − rSₙ = a − arⁿ, so
Sₙ = a(1 − rⁿ)/(1 − r) (r ≠ 1).
If |r| < 1 then rⁿ → 0 and the series converges to a/(1 − r), with remainder arⁿ/(1 − r) after n terms. If |r| ≥ 1 the terms do not shrink to 0, so no sum is possible. The square of halves is the case a = r = 1/2: the sum is (1/2)/(1 − 1/2) = 1.
Zeno of Elea, in the fifth century BC, argued that a runner can never finish a race, since first they must reach the halfway point, then half the remaining distance, and so on without end. His arguments survive through Aristotle’s Physics. The modern reply is that infinitely many stages can take a finite total time, because 1/2 + 1/4 + 1/8 + … converges. Whether that fully answers Zeno is still debated by philosophers, but it settles the arithmetic.
Press Build and stack up to 50 blocks at the edge of a table. Each one overhangs the next by a harmonic fraction, and the top block drifts as far out as you like.
The blocks are idealised: rigid, identical, uniform, one per level, and each balance point lies exactly on the edge below, so this is the limit of what stays up. Blocks are drawn squashed vertically so that 50 of them fit.
Build from the top. One block balances with its centre on the edge below, so it overhangs by 1/2. Two blocks have their common centre of mass a quarter of a block to the right of the second block’s centre, and that point must sit over the edge of the third block: the second block overhangs by 1/4. In general the top k blocks have total mass k, and sliding them so their centre of mass sits on the edge below makes block k overhang its support by 1/(2k). The total is
overhang = ½(1 + 1/2 + 1/3 + … + 1/N) = H_N / 2.
The harmonic numbers grow without bound. Nicole Oresme proved it around 1350 by grouping: 1/3 + 1/4 > 1/2, 1/5 + … + 1/8 > 1/2, and so on, so H₂ₖ > 1 + k/2. The picture on the right gives a sharper bound: the rectangle of height 1/k on [k, k + 1] lies above the curve 1/x, so H_N > ∫₁ᴺ⁺¹ dx/x = ln(N + 1). In fact H_N − ln N tends to Euler’s constant γ ≈ 0.5772, so 4 blocks pass one block-length of overhang, 31 pass two, 227 pass three, and around 272 million are needed for ten.
One block per level is far from the best design. Mike Paterson and Uri Zwick showed in 2006 that if several blocks may share a level and act as counterweights, the overhang can grow like N^(1/3), far faster than ½ ln N, and a follow-up paper with Yuval Peres, Mikkel Thorup and Peter Winkler proved that order N^(1/3) is the best possible.
Choose a series and watch a million partial sums, then read what each test says. Slide p through 1 on ∑ 1/nᵖ and see which tests notice.
aₙ → 0. That is necessary for convergence but not enough, so this test says nothing.
|aₙ₊₁/aₙ| → 1. The ratio test cannot tell, as for every p-series.
|aₙ|^(1/n) → 1. Inconclusive: polynomial-sized terms always give 1.
∫₁^∞ dx/x² = 1. The terms come from a positive, decreasing f, so the sum converges with the integral.
|aₙ|/bₙ → 1: a finite positive limit, so the series converges along with ∑ 1/n².
The terms do not alternate in sign.
Benchmark for the comparison test
Partial sums are added term by term in double precision up to N = 10⁶ and plotted at about 60 points per decade; a divergent geometric series stops once it passes 10⁹. Test verdicts use each series’ exact asymptotic form (for n!/nⁿ, Stirling’s formula), not the plotted numbers.
No finite computation can prove that a series converges: the partial sums of ∑ 1/(n ln n) are below 4 after a million terms and still go to infinity. So we compare with series we understand.
Euler found in 1734 that ∑ 1/n² = π²/6, solving the Basel problem that had defeated Jakob Bernoulli. Most series have no such closed form: nobody knows a simple formula for ∑ 1/n³, even though it certainly converges.
Choose any target T, even π, and watch the terms of 1 − 1/2 + 1/3 − … reshuffled so that the sum lands on it. Below, an absolutely convergent series ignores the reshuffle.
The greedy rule takes a positive term when the sum is at most T and a negative term when it is above. With positives and negatives in the fixed ratio p : q, the rearranged sum is ln 2 + ½ ln(p/q), which is where e^(2T)/4 comes from. Sums are plotted at most 4000 points at a time.
The alternating harmonic series 1 − 1/2 + 1/3 − 1/4 + … converges to ln 2 by the alternating series test, but the series of absolute values is the harmonic series, which diverges. Such a series is called conditionally convergent, and its positive part and negative part each diverge on their own. That is what lets the greedy rule work: there is always enough positive mass left to climb above any T and enough negative mass to fall below it, while the terms themselves shrink to 0, so the swings around T die out.
Bernhard Riemann proved in 1854 (published 1867) that a conditionally convergent series of real numbers can be rearranged to converge to any real number, or to diverge. Dirichlet had shown in 1837 that an absolutely convergent series gives the same sum in every order, which is what the lower strip shows for 1 − 1/4 + 1/9 − … = π²/12.
For the alternating harmonic series there is even a formula: taking p positive terms for every q negative ones gives ln 2 + ½ ln(p/q). Two positives per negative gives (3/2) ln 2, a classical example.
Spot a mistake or have an idea?