Math paradoxes

Glossary / Harmonic series

Glossary · harmonic series

How far can a stack of books hang over a table edge?

Put a book on the edge of a table and slide it out until it is about to fall: it sticks out by exactly half. Now pile books on top of one another, each a bit further out than the one below. It seems you can't get far this way: the stack will topple as soon as the top book is a length or two past the edge. In fact, with enough books, the top one can hang out over the floor as far as you like. It is all down to the harmonic series: the sum 1 + ½ + ⅓ + ¼ + …, which grows without end, if unbelievably slowly.

In plain wordspictures, analogies, small numbers

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The book stack paradox

Add books. Each new one goes underneath, and the whole stack above it is pushed out as far as it can go without falling. The scale stays fixed: the dashed lines are book lengths past the table edge. How many books does it take for the top one to be completely past the edge?

Books–
Overhang, in book lengths–

How many books it takes for the top one to overhang by:

    The picture holds up to 300 books; the rest we calculated.

    Ever slower, but without end

    The first book sticks out by half a length, the second adds a quarter, the third a sixth, the fourth an eighth. Each next one gives less, and it feels as if such crumbs must stop somewhere. But they don't. With four books the top one is already entirely over the floor, with 31 it is two lengths out. The price grows terribly: five lengths take 12,367 books. At 3 cm each, that is a stack 371 m tall, higher than the Eiffel Tower. And yet there is no limit: any distance can be exceeded.

    Rigoroustables, charts, formulas

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    The sum of reciprocals

    The harmonic series is the sum of the reciprocals of the whole numbers: 1 + ½ + ⅓ + ¼ + … The sum of the first n terms is called the n-th harmonic number and written Hn. The series diverges: Hn grows without bound, even though the terms shrink to zero.

    H4 = 1 + 1/2 + 1/3 + 1/4 = 25/12 ≈ 2.08 · H10 ≈ 2.93 · H100 ≈ 5.19 · H1000 ≈ 7.49

    The book stack is the same series, just halved: n books overhang by ½ · Hn book lengths. With 4 books that is ½ · 25/12 = 25/24, a little over one length.

    A series that stops and one that doesn't

    Compare it with 1 + ½ + ¼ + ⅛ + …, where each term is half the one before. Its terms also shrink to zero, even faster. But it never gets past 2: each step covers only half of the distance still left to two. The harmonic series barely trails it at first, then climbs slowly but relentlessly. Switch the scale.

    Harmonic, last sum–
    Halving, last sum–

    How painfully slow

    How many terms does it take for the sum to pass a given number? Add one term a second and see how long it would take.

    Terms needed to pass it–
    At one term a second–

    What to take from this

    • Small terms don't mean a finite sumTerms shrinking to zero is necessary for a sum to settle, but not enough. The harmonic series is the classic counterexample.
    • What matters is how fast they shrink1/n² already gives a finite sum, π²/6 ≈ 1.645. Halvings 1/2n all the more. And 1/n is the borderline rate at which the sum still runs off to infinity.
    • A computer won't noticeAdd a million terms and you get about 14.4. A billion, about 21.3. It looks like a number the sum is creeping towards, but there is none. Divergence is proved, not observed.
    • It grows like a logarithmTen times more terms adds only about 2.3 to the sum. That is why harmonic numbers turn up wherever something "slowly piles up" with each new step.
    Deepergeneralisations, theory, history

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    Oresme's proof: grouping the terms

    Around 1350 the French philosopher Nicole Oresme spotted a simple trick. Group the terms: the second group has two terms, the third four, the fourth eight. In each group every term is at least as big as the last one, so the whole group is at least a half.

    1 + ½ + (⅓ + ¼) + (1/5 + 1/6 + 1/7 + 1/8) + … ≥ 1 + ½ + (¼ + ¼) + (⅛ + ⅛ + ⅛ + ⅛) + … = 1 + ½ + ½ + ½ + …

    H2k ≥ 1 + k / 2

    There are infinitely many halves, so the sum will exceed any number. See how the bound holds from below.

    Terms 2k–
    H2k vs the bound 1 + k/2–

    The bound is crude: the real sum grows faster than ½ per group. But that is enough for a proof.

    Hn ≈ ln n + γ

    The sum 1/1 + 1/2 + … + 1/n is almost the area under the curve 1/x from 1 to n, and that area is the natural logarithm ln n. The gap between them doesn't grow but settles on a constant γ ≈ 0.5772, named after Leonhard Euler. It is still unknown whether γ can even be written as an ordinary fraction.

    Hn = ln n + γ + 1/(2n) − …

    Exact Hn–
    ln n + γ–
    Difference–
    Correction 1/(2n)–

    This gives the numbers from the section above: for Hn to pass t you need n ≈ et − γ. For 10 that is about 12,367; for 100, about 1.5 · 1043 terms. That is exactly how many books it takes for a stack to overhang by 50 lengths.

    Why the overhang is exactly ½ · Hn

    A stack stands as long as the common centre of mass of every top part lies over the book beneath it. So the furthest you can go is to place the centre of mass of the top k books exactly over the edge of book k + 1. Suppose that is so, and measure from that edge. The centre of the k books is at the edge, at zero, and the centre of book k + 1 is half a length further in. The common centre of the k + 1 books is the weighted average:

    (k · 0 + 1 · ½) / (k + 1) = 1 / (2(k + 1))

    That is exactly where the edge of the next book must be, so book k + 1 sticks out past it by 1/(2(k + 1)). Adding the shifts 1/2, 1/4, 1/6, …, 1/(2n) gives ½ · Hn. The last shift is over the table: the centre of mass of the whole stack sits exactly on the table edge.

    That is the record for a stack with one book on each level. If several books may lie side by side and act as counterweights, the overhang grows much faster: Mike Paterson and Uri Zwick showed this in 2006.

    Strike out the nines and the sum stops

    One more paradox. Remove from the harmonic series every term 1/n whose n contains the digit 9: 1/9, 1/19, 1/90, 1/199… It seems we threw out a trifle, as nines are rare among small numbers. But the sum becomes finite, below 80. There are 8 · 9d − 1 d-digit numbers without a nine, and each of their reciprocals is at most 1/10d − 1. So the whole sum is at most

    8 · (1 + 9/10 + (9/10)2 + …) = 8 · 10 = 80

    The point is that among very long numbers almost all contain a nine. We are not throwing out a trifle but nearly the whole series. This series was studied by Aubrey Kempner in 1914.

    Where it shows up

    Other terms