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Sums that blow up

Infinity and strange math · level 4, lesson 3 of 6

A question to start

You have a pile of identical books and a table. Stack the books one on top of another, near the edge of the table, so that each book sticks out a little past the one below it. How far out past the edge of the table can the top book reach, before the pile falls over?

Most people guess: "Less than one book length. The top book could never be completely past the edge." In fact, with just 4 books the top book can be completely beyond the table. And with enough books, you can reach as far out as you like: 2 book lengths, 10, a hundred. The reason is one of the most surprising sums in mathematics.

The idea

The harmonic series

In the last two lessons, endless sums like \(\frac12 + \frac14 + \frac18 + \cdots\) settled on a finite total. Now look at this one:

\[1 + \frac12 + \frac13 + \frac14 + \frac15 + \frac16 + \cdots\]

Each piece is 1 divided by the next counting number. This is called the harmonic series. (The name comes from music: a string vibrating in halves, thirds, quarters, ... gives the notes called harmonics.)

The pieces get smaller and smaller, shrinking towards 0, just like the halving sum. So you might expect the total to settle down. Let us look at partial sums.

Number of pieces 1 2 4 10 100 1000 1,000,000
Partial sum (about) 1 1.5 2.08 2.93 5.19 7.49 14.39

It grows, but very slowly. A million pieces only get you to about 14.4. It looks as if it might creep up towards some limit, maybe 20 or 30. But it does not. The harmonic series grows past every number: past 100, past a million, past anything you name. We say it diverges, or "blows up".

How can anyone know that, when the numbers grow so slowly that no computer could ever add enough pieces to reach 100? With a proof.

The grouping proof

This proof is about 650 years old. It was found by the French thinker Nicole Oresme, around 1350. The idea is to put the pieces into groups, where each group adds up to at least one half.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 1/2 bar k has height 1/k; each coloured group adds up to at least 1/2

Write the sum like this, with brackets:

\[1 + \frac12 + \left(\frac13 + \frac14\right) + \left(\frac15 + \frac16 + \frac17 + \frac18\right) + \left(\frac19 + \cdots + \frac1{16}\right) + \cdots\]

Each group ends at a power of 2 (4, 8, 16, 32, ...), and each group has twice as many pieces as the one before.

  • Group \(\frac13 + \frac14\). Both pieces are at least \(\frac14\) (since \(\frac13\) is bigger than \(\frac14\)). Two pieces, each at least \(\frac14\): the group is at least \(2 \times \frac14 = \frac12\).
  • Group \(\frac15 + \frac16 + \frac17 + \frac18\). All four are at least \(\frac18\). So the group is at least \(4 \times \frac18 = \frac12\).
  • Group \(\frac19\) to \(\frac1{16}\). Eight pieces, each at least \(\frac1{16}\): at least \(8 \times \frac1{16} = \frac12\).
  • Every group after that has twice as many pieces, each at least half as big, so it is always at least \(\frac12\).

So the sum is at least

\[1 + \frac12 + \frac12 + \frac12 + \frac12 + \cdots\]

and adding \(\frac12\) over and over grows past any number. Two halves make 1, so 200 groups make at least 100, and 2,000,000 groups make at least a million. The harmonic series has no finite total.

We can even say how many pieces are enough. Up to the piece \(\frac1{2^k}\), there are the 1, then \(k\) groups (the \(\frac12\) counts as the first), each at least \(\frac12\). So

\[1 + \frac12 + \frac13 + \cdots + \frac1{2^k} \ge 1 + \frac{k}2.\]

With \(k = 4\) (up to \(\frac1{16}\)), the sum is at least 3. The table agrees: the true value up to \(\frac1{16}\) is about 3.38.

Shrinking pieces are not enough

Here is the big lesson. For an endless sum to have a finite total, the pieces must shrink towards 0 (if they did not, you would keep adding chunks of the same size, and the total would grow forever). But shrinking is not enough. The pieces must shrink fast enough.

Compare the two sums piece by piece.

Piece number 1 2 3 4 5 8 16
Halving sum \(\frac12\) \(\frac14\) \(\frac18\) \(\frac1{16}\) \(\frac1{32}\) \(\frac1{256}\) \(\frac1{65536}\)
Harmonic sum 1 \(\frac12\) \(\frac13\) \(\frac14\) \(\frac15\) \(\frac18\) \(\frac1{16}\)

The halving pieces shrink by half every single step, so they quickly become tiny. The harmonic pieces shrink much more slowly: the 16th is still \(\frac1{16}\). Those slow pieces pile up into groups worth \(\frac12\) again and again.

Stacking books

Now back to the books. Make each book 1 unit long. The trick is to build the pile from the top down.

The physics fact we need is about balance. Every object has a balance point (scientists call it the centre of mass): the point where you could balance it on a fingertip. For a book, it is the middle. A pile of books stays up as long as the balance point of the books above each edge is not past that edge.

  • The top book can stick out until its middle is right over the edge of the book below: it sticks out \(\frac12\).
  • The top two books together: their balance point is \(\frac14\) in from the right end of the second book. (The top book's middle is at the second book's edge, and the second book's middle is \(\frac12\) in from its edge. Two equal weights balance halfway between, \(\frac14\) in.) So the second book can stick out \(\frac14\) past the third.
  • The top three books: their balance point is \(\frac16\) in from the end of the third book, so the third can stick out \(\frac16\) past the fourth.
  • In general, the \(k\)th book from the top can stick out \(\frac1{2k}\) past the one below.
table 1/2 1/4 1/6 1/8 1/10 one book length each book sticks out 1/2, 1/4, 1/6, 1/8, 1/10 past the one below

Why \(\frac1{2k}\)? The top \(k - 1\) books have their balance point right over the edge of book \(k\). Book \(k\) itself has its balance point \(\frac12\) in from that edge. So we have \(k - 1\) books' worth of weight at the edge and 1 book's weight \(\frac12\) in. The balance point of all \(k\) is the average position, weighted by the number of books:

\[\frac{(k - 1) \times 0 + 1 \times \frac12}{k} = \frac1{2k}\]

in from the edge of book \(k\). That is how far book \(k\) can stick out past book \(k + 1\) (or past the table).

So with \(n\) books the top one reaches past the table by

\[\frac12 + \frac14 + \frac16 + \cdots + \frac1{2n} = \frac12\left(1 + \frac12 + \frac13 + \cdots + \frac1n\right),\]

which is half of the harmonic series. Since the harmonic series grows past every number, so does half of it. With enough books, the overhang can be as large as you like!

How many books? With 4 books the overhang is \(\frac12 \times \left(1 + \frac12 + \frac13 + \frac14\right) = \frac12 \times \frac{25}{12} = \frac{25}{24}\), a little more than 1. So the top book is completely past the edge. To reach 2 book lengths you need 31 books; to reach 3 you need 227. The harmonic series grows slowly, so the stack gets tall very quickly. (Real books are not perfectly even, so in practice you leave a little safety margin.)

Worked examples

1. The first few partial sums exactly. Find \(1 + \frac12 + \frac13 + \frac14\).

Use twelfths: \(\frac{12}{12} + \frac6{12} + \frac4{12} + \frac3{12} = \frac{25}{12}\), which is about 2.08.

2. Checking one group. Show that \(\frac15 + \frac16 + \frac17 + \frac18 \ge \frac12\), and find how much more it is.

  • Each of the four pieces is at least \(\frac18\), so the sum is at least \(\frac48 = \frac12\).
  • Exactly: with denominator 840, the sum is \(\frac{168 + 140 + 120 + 105}{840} = \frac{533}{840}\), about 0.635. So it is about 0.135 more than \(\frac12\).

3. Guaranteeing a total of 5. How many pieces of the harmonic series are certainly enough to pass 5?

  • Up to \(\frac1{2^k}\) the sum is at least \(1 + \frac k2\). We need \(1 + \frac k2 \ge 5\), so \(\frac k2 \ge 4\), so \(k = 8\).
  • \(2^8 = 256\) pieces are enough. (The proof gives a safe answer, not the smallest one. In fact the sum first passes 5 at the 83rd piece.)

4. A sum that looks small but blows up. Does \(\frac1{10} + \frac1{20} + \frac1{30} + \frac1{40} + \cdots\) have a finite total?

  • Every piece is \(\frac1{10}\) of a harmonic piece: \(\frac1{20} = \frac1{10} \times \frac12\).
  • So the sum is \(\frac1{10}\) of the harmonic series. A tenth of something that grows past every number still grows past every number (to pass 100, wait until the harmonic series passes 1000).
  • No finite total: it blows up.

5. Three books. How far past the table does the top of a 3-book stack reach?

\(\frac12 + \frac14 + \frac16 = \frac6{12} + \frac3{12} + \frac2{12} = \frac{11}{12}\) of a book length. Just short of a whole book. The 4th book is what gets you past 1.

Try it

  1. Find \(1 + \frac12 + \frac13 + \frac14 + \frac15 + \frac16\) as a fraction.
  2. Show that \(\frac19 + \frac1{10} + \cdots + \frac1{16}\) (eight pieces) is at least \(\frac12\).
  3. Using the grouping proof, the sum up to \(\frac1{1024}\) is at least what? (\(1024 = 2^{10}\).)
  4. How far past the table does a 2-book stack reach? A 5-book stack?
  5. Does \(\frac12 + \frac14 + \frac16 + \frac18 + \cdots\) (one over each even number) have a finite total? Explain.
  6. Does \(1 + 1 + 1 + \cdots\) have a finite total? Does \(1 - 1 + 1 - 1 + \cdots\)?
  7. (Harder) Without the grouping proof, here is another argument. Suppose the harmonic series had a finite total \(H\). Use the answer to question 5, together with the odd pieces \(1 + \frac13 + \frac15 + \cdots\), to reach a contradiction. (Hint: compare each odd piece with the even piece after it.)
Answers
  1. With denominator 60: \(\frac{60 + 30 + 20 + 15 + 12 + 10}{60} = \frac{147}{60} = \frac{49}{20} = 2.45\).
  2. Each of the eight pieces is at least \(\frac1{16}\), so the sum is at least \(\frac8{16} = \frac12\).
  3. At least \(1 + \frac{10}2 = 6\). (The true value is about 7.5.)
  4. Two books: \(\frac12 + \frac14 = \frac34\). Five books: \(\frac12 + \frac14 + \frac16 + \frac18 + \frac1{10}\). With denominator 120: \(\frac{60 + 30 + 20 + 15 + 12}{120} = \frac{137}{120}\), about 1.14 book lengths.
  5. No. Each piece is half of a harmonic piece (\(\frac1{2k} = \frac12 \times \frac1k\)), so the sum is half the harmonic series, which grows past every number. It is exactly the book-stack sum.
  6. \(1 + 1 + 1 + \cdots\) has partial sums 1, 2, 3, ... which grow forever: no finite total. \(1 - 1 + 1 - 1 + \cdots\) has partial sums 1, 0, 1, 0, ... which never settle on one number, so it has no total either. (Pieces that do not shrink to 0 never give a total.)
  7. The even pieces add up to \(\frac12 H\) (question 5 shows they are half the harmonic series). Since all the pieces add to \(H\), the odd pieces \(1 + \frac13 + \frac15 + \cdots\) must add up to the other half, \(\frac12 H\). But each odd piece is bigger than the even piece after it: \(1 > \frac12\), \(\frac13 > \frac14\), \(\frac15 > \frac16\), and so on. So the odd pieces add to more than the even pieces: more than \(\frac12 H\), in fact at least \(\frac12\) more, since \(1 - \frac12 = \frac12\) already. Both cannot be \(\frac12 H\). The contradiction shows there is no finite total \(H\).

Common mistake

"The pieces go to zero, so the sum must be finite." This is the most common mistake about endless sums, and the harmonic series is the famous counterexample (an example that shows a claim is false). Shrinking to zero is needed, but it is not enough.

Trusting a computer's partial sums. After a million pieces the sum is only about 14.4, and after a billion pieces about 21. A computer would make it look as if the sum levels off. Only a proof can show it never does.

Where it leads