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Infinite geometric series

Patterns and sequences · level 7, lesson 3 of 6

A question to start

A ball is dropped from a height of 10 metres. Each time it bounces, it rises to 60% of the height it fell from: 6 m, then 3.6 m, then 2.16 m, and so on. In theory it bounces forever, an infinite number of times.

How far does it travel altogether, up and down? Infinitely far, because it makes infinitely many bounces? Or some ordinary number of metres? The answer is exactly 40 metres, and by the end of this lesson you will be able to prove it.

The idea

Adding infinitely many numbers

In level 6 you added the first \(n\) terms of a geometric sequence with first term \(a\) and common ratio \(r\):

\[S_n = a + ar + ar^2 + \cdots + ar^{n-1} = \frac{a(1 - r^n)}{1 - r} \qquad (r \ne 1).\]

What could it mean to add all the terms, forever? Nobody can do infinitely many additions. Instead we look at the partial sums \(S_1, S_2, S_3, \ldots\), the totals so far, and ask whether they settle down towards one number. If they do, that number is called the sum to infinity and written \(S_\infty\) or \(S\). If they do not settle, we say the series has no sum (it diverges).

A square of paper

Take a square of area 1. Shade half of it. Then shade half of what is left, then half of what is still left, and so on.

1/2 1/4 1/8 1/16 The pieces fill the square more and more; the gap left after n pieces is (1/2)ⁿ.

After one piece, \(\frac12\) is shaded; after two, \(\frac34\); after three, \(\frac78\):

pieces \(n\) 1 2 3 4 5 6 10
shaded \(S_n\) 0.5 0.75 0.875 0.9375 0.96875 0.984375 0.9990234375
gap \(1 - S_n\) 0.5 0.25 0.125 0.0625 0.03125 0.015625 0.0009765625

The pieces never overlap and never spill out of the square, so the shaded area never passes 1. And the gap left over halves every time, so it becomes smaller than any size you name: smaller than a thousandth after 10 pieces, smaller than a millionth after 20. The partial sums close in on 1, and nothing less than 1 can be the number they settle on. So

\[\frac12 + \frac14 + \frac18 + \frac1{16} + \cdots = 1.\]

The general case: watch the remainder

Now any geometric series. Split the formula for \(S_n\) into two pieces:

\[S_n = \frac{a(1 - r^n)}{1 - r} = \frac{a}{1 - r} - \frac{a\,r^n}{1 - r}.\]

The first piece, \(\frac{a}{1 - r}\), does not depend on \(n\) at all. The second piece is the remainder: how far the partial sum is from \(\frac{a}{1 - r}\). Everything depends on what \(r^n\) does as \(n\) grows.

  • If \(-1 < r < 1\) (written \(|r| < 1\)), then \(r^n\) shrinks towards 0. For example \(0.6^{10} \approx 0.006\) and \(0.6^{30} \approx 0.000\,000\,2\). So the remainder shrinks to 0 and the partial sums settle on \(\frac{a}{1-r}\).

This gives the main result of the lesson:

\[S_\infty = a + ar + ar^2 + \cdots = \frac{a}{1 - r} \qquad \text{when } |r| < 1.\]

Why does \(r^n\) shrink to 0?

It feels obvious that multiplying by 0.99 again and again gets you close to 0, but 0.99 is very close to 1, so it is worth a real reason. Take \(0 < r < 1\) and write \(\frac{1}{r} = 1 + h\), where \(h > 0\). (For \(r = 0.99\), \(h\) is about 0.0101.) Multiplying out \((1 + h)^n\) gives \(1 + nh\) plus other terms that are all positive, so

\[(1 + h)^n \ge 1 + nh, \qquad\text{and therefore}\qquad r^n = \frac{1}{(1 + h)^n} \le \frac{1}{1 + nh}.\]

As \(n\) grows, \(1 + nh\) grows past any number you like (it goes up by \(h\) each time), so \(\frac{1}{1 + nh}\), and \(r^n\) with it, gets as close to 0 as you like. For a negative \(r\) between \(-1\) and 0, \(r^n\) has the same size as \(|r|^n\), only with flipping signs, so it shrinks to 0 too. (The inequality \((1 + h)^n \ge 1 + nh\) is called Bernoulli's inequality; in level 8 you will prove it by induction.)

When there is no sum

  • If \(r = 1\): \(a + a + a + \cdots\). The partial sums are \(a, 2a, 3a, \ldots\), which grow without end (if \(a \ne 0\)).
  • If \(r > 1\): the terms themselves grow, so the partial sums grow without end. \(1 + 2 + 4 + 8 + \cdots\) has no sum.
  • If \(r = -1\): \(1 - 1 + 1 - 1 + \cdots\) has partial sums \(1, 0, 1, 0, \ldots\), which jump for ever and never settle.
  • If \(r < -1\): the terms swing further and further each way, and the partial sums swing wider and wider.

In every case with \(|r| \ge 1\) the terms do not shrink to 0, and a series whose terms do not shrink to 0 can never settle. The formula \(\frac{a}{1 - r}\) still gives a number when you put in \(r = 2\), but that number (\(-a\)) means nothing. Always check \(|r| < 1\) first.

Worked examples

1. A first sum. Find \(8 + 4 + 2 + 1 + \cdots\).

\(a = 8\), \(r = \frac12\), and \(|r| < 1\), so \(S_\infty = \frac{8}{1 - \frac12} = \frac{8}{\frac12} = 16\).

2. Alternating signs. Find \(1 - \frac12 + \frac14 - \frac18 + \cdots\).

\(a = 1\), \(r = -\frac12\), and \(|r| < 1\), so \(S_\infty = \frac{1}{1 - (-\frac12)} = \frac{1}{\frac32} = \frac23\). The partial sums \(1, 0.5, 0.75, 0.625, 0.6875, \ldots\) hop above and below \(\frac23\), closer each time.

3. A repeating decimal as a fraction. Write \(0.272727\ldots\) as a fraction.

\[0.272727\ldots = \frac{27}{100} + \frac{27}{10\,000} + \frac{27}{1\,000\,000} + \cdots,\]

with \(a = \frac{27}{100}\) and \(r = \frac{1}{100}\). So the sum is \(\frac{27/100}{99/100} = \frac{27}{99} = \frac{3}{11}\). Check: \(3 \div 11 = 0.2727\ldots\) ✓. Every repeating decimal is a geometric series, which is why every repeating decimal is a fraction. (What this says about \(0.999\ldots\) is a puzzle for Infinity.)

4. The bouncing ball. Back to the ball dropped from 10 m that rebounds to 60% each time.

10 6 3.6 2.16 1.3
  • First, the drop: 10 m down.
  • After that, every bounce goes up and then down the same height. The bounce heights \(6, 3.6, 2.16, \ldots\) form a geometric series with \(a = 6\) and \(r = 0.6\), whose sum is \(\frac{6}{1 - 0.6} = \frac{6}{0.4} = 15\) m.
  • Up and down doubles this: 30 m.

Total: \(10 + 30 = 40\) metres. Infinitely many bounces, finite distance, because the bounces shrink fast enough. (Real balls stop after a few dozen bounces, but the model is very close.)

5. A snowflake with finite area and endless edge. Start with an equilateral triangle of area \(A\). On the middle third of every side, stick on a small equilateral triangle pointing outwards. Repeat on every side of the new shape, forever. This is the Koch snowflake.

  • Step 1 adds 3 triangles, each \(\frac19\) of the area of the big one (a third the side length gives a ninth the area): \(\frac{3A}{9} = \frac{A}{3}\).
  • Each step multiplies the number of sides by 4 and shrinks the new triangles' area by 9, so each step adds \(\frac49\) of the area the step before added: \(\frac{A}{3}, \frac{A}{3} \cdot \frac49, \ldots\)

The extra area is a geometric series with \(r = \frac49\): \(\frac{A/3}{1 - 4/9} = \frac{A}{3} \cdot \frac95 = \frac{3A}{5}\). The snowflake's area is \(A + \frac{3A}{5} = \frac{8A}{5}\), finite. But each step replaces every side by 4 pieces each a third as long, multiplying the perimeter by \(\frac43\). With \(r = \frac43 > 1\), the perimeter grows without end. The same picture holds one series with a sum and one without.

6. How many terms are enough? How many terms of \(1 + \frac12 + \frac14 + \cdots\) get within 0.001 of the sum, 2?

The remainder after \(n\) terms is \(\frac{a r^n}{1 - r} = \frac{(1/2)^n}{1/2} = \frac{1}{2^{\,n-1}}\). We need \(2^{\,n-1} > 1000\). Since \(2^9 = 512\) and \(2^{10} = 1024\), we need \(n - 1 \ge 10\): 11 terms.

Try it

  1. Find \(12 + 6 + 3 + \cdots\).
  2. Find \(5 - \frac53 + \frac59 - \cdots\).
  3. Write \(0.363636\ldots\) as a fraction in lowest terms.
  4. Which of these have a sum? Find it where it exists. (a) \(1 + 1.1 + 1.21 + \cdots\) (b) \(9 - 6 + 4 - \cdots\) (c) \(2 - 2 + 2 - 2 + \cdots\)
  5. For which \(x\) does \(1 + x + x^2 + x^3 + \cdots\) have a sum? Find \(x\) if the sum is 5.
  6. A ball is dropped from 3 m and each bounce reaches \(\frac23\) of the height before. Find the total distance.
  7. A geometric series has sum 4, and the series of the squares of its terms has sum \(\frac{16}{3}\). Find \(a\) and \(r\).
  8. How many terms of \(1 + \frac13 + \frac19 + \cdots\) are needed for the partial sum to be within 0.0001 of the sum?
Answers
  1. \(a = 12\), \(r = \frac12\): \(\frac{12}{1/2} =\) 24.
  2. \(a = 5\), \(r = -\frac13\): \(\frac{5}{1 + 1/3} = \frac{5}{4/3} =\) \(\frac{15}{4}\).
  3. \(a = \frac{36}{100}\), \(r = \frac{1}{100}\): \(\frac{36}{99} =\) \(\frac{4}{11}\).
  4. (a) \(r = 1.1 > 1\): no sum. (b) \(r = -\frac23\), \(|r| < 1\): \(\frac{9}{1 + 2/3} = \frac{9}{5/3} =\) \(\frac{27}{5}\). (c) \(r = -1\): no sum (partial sums \(2, 0, 2, 0, \ldots\)).
  5. It is geometric with \(a = 1\) and \(r = x\), so it has a sum exactly when \(-1 < x < 1\), and then the sum is \(\frac{1}{1 - x}\). Setting \(\frac{1}{1 - x} = 5\) gives \(1 - x = \frac15\), so \(x = \frac45\).
  6. Drop 3 m. Bounces: \(2 + \frac43 + \cdots\) with sum \(\frac{2}{1 - 2/3} = 6\) m, doubled for up and down: 12 m. Total 15 m.
  7. \(\frac{a}{1 - r} = 4\) gives \(a = 4(1 - r)\). The squares \(a^2, a^2r^2, a^2r^4, \ldots\) form a geometric series with ratio \(r^2\), sum \(\frac{a^2}{1 - r^2}\). Substituting, \(\frac{16(1 - r)^2}{(1 - r)(1 + r)} = \frac{16(1 - r)}{1 + r} = \frac{16}{3}\), so \(3(1 - r) = 1 + r\), \(r = \frac12\), and \(a = 2\). (Check: \(2 + 1 + \frac12 + \cdots = 4\) and \(4 + 1 + \frac14 + \cdots = \frac{4}{3/4} = \frac{16}{3}\). ✓)
  8. The sum is \(\frac{1}{1 - 1/3} = \frac32\). The remainder is \(\frac{(1/3)^n}{2/3} = \frac{3}{2 \cdot 3^n}\). We need \(\frac{3}{2 \cdot 3^n} < 0.0001\), so \(3^n > 15\,000\). \(3^8 = 6561\) is too small, \(3^9 = 19\,683\) is enough: 9 terms.

Common mistake

Using \(\frac{a}{1-r}\) when \(|r| \ge 1\). For \(1 + 3 + 9 + \cdots\) the formula gives \(\frac{1}{1 - 3} = -\frac12\), a negative "sum" of positive numbers. That is a sure sign the formula did not apply.

Forgetting the extra trip in bounce questions. Every bounce after the first drop is travelled twice (up, then down), but the first drop is travelled once. Add the drop separately.

Where it leads