Some geometric series have infinitely many terms but still add up to a finite number. This note covers when an infinite geometric series converges, the formula for its sum, and how it turns recurring decimals into fractions. It also explains situations such as a bouncing ball that travels a finite total distance.
Key ideas
A geometric series with first term \(u_1\) and common ratio \(r\) has partial sums \(S_n = \dfrac{u_1(1 - r^n)}{1 - r}\). The key question is what happens as \(n\) gets very large.
- If \(|r| \lt 1\), the term \(r^n\) gets closer and closer to 0 as \(n\) grows. For example, \(0.5^{10} \approx 0.001\) and \(0.5^{50}\) is almost zero. The partial sums settle towards a fixed value. We say the series converges.
- If \(|r| \ge 1\), the term \(r^n\) does not shrink to zero. The sums grow without limit, or keep jumping around. We say the series diverges and has no sum to infinity.
The condition \(|r| \lt 1\) means \(-1 \lt r \lt 1\). It includes negative ratios, for which the partial sums swing above and below the limit while getting closer to it.
When the series converges, letting \(r^n \to 0\) in the formula for \(S_n\) gives the sum to infinity:
\[S_\infty = \frac{u_1}{1 - r}, \qquad |r| \lt 1\]
Recurring decimals. A recurring decimal is an infinite geometric series in disguise. For example, \(0.\overline{36} = 0.36 + 0.0036 + 0.000036 + \dots\) has \(u_1 = 0.36\) and \(r = 0.01\). If a decimal has a repeating block of \(k\) digits, the ratio is \(10^{-k}\). If the decimal also has digits before the repeating block starts, split off the non-repeating part and apply the formula to the repeating part.
Conditions on a variable. If the common ratio contains an unknown, such as \(r = \dfrac{x}{4}\), solve the inequality \(|r| \lt 1\) to find the values of \(x\) for which the series converges. For \(|x/4| \lt 1\) we get \(-4 \lt x \lt 4\).
Modelling. When something repeatedly changes by a fixed fraction, such as a ball losing part of its height on each bounce, the total quantity is often an infinite geometric series. Identify the first term and the ratio carefully. Be careful when the quantity has a different starting pattern, such as a ball that first falls once and then goes up and down each time. In that case, treat the first drop separately.
Sometimes you are given the sum to infinity and must find \(u_1\) or \(r\). Substitute what you know into \(S_\infty = \dfrac{u_1}{1-r}\) and solve for the unknown. Check that the ratio you get satisfies \(|r| \lt 1\).
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| \(|r| \lt 1\) | Condition for an infinite geometric series to converge. |
| \(S_\infty = \dfrac{u_1}{1 - r}\) | Sum to infinity of a convergent geometric series. |
| \(S_n = \dfrac{u_1(1 - r^n)}{1 - r}\) | Sum of the first \(n\) terms, from which \(S_\infty\) arises when \(r^n \to 0\). |
| Recurring decimal | A decimal whose digits repeat; it equals an infinite geometric series with \(r = 10^{-k}\). |
| Convergent / divergent | A series with a finite sum to infinity / one without. |
Worked examples
Example 1 Write \(0.\overline{36} = 0.363636\ldots\) as a fraction in its simplest form.
Write the decimal as a series: \(0.36 + 0.0036 + 0.000036 + \dots\)
This is geometric with \(u_1 = 0.36\) and \(r = \dfrac{0.0036}{0.36} = 0.01\). Since \(|r| \lt 1\), the sum to infinity exists.
\(S_\infty = \dfrac{0.36}{1 - 0.01} = \dfrac{0.36}{0.99} = \dfrac{36}{99}\).
Divide the top and bottom by 9: \(\dfrac{36}{99} = \dfrac{4}{11}\).
Answer: \(\dfrac{4}{11}\).
Example 2 A ball is dropped from a height of 10 m. Each time it hits the ground it rebounds to 60% of the height it fell from. Find the total distance the ball travels before it comes to rest.
The first drop is 10 m. After that, each rebound involves an upward and a downward journey of the same height.
The first rebound height is \(10 \times 0.6 = 6\) m, so the first up-and-down distance is \(2 \times 6 = 12\) m. The next is \(12 \times 0.6 = 7.2\) m, and so on.
The up-and-down distances form a geometric series with \(u_1 = 12\) and \(r = 0.6\).
\(S_\infty = \dfrac{12}{1 - 0.6} = \dfrac{12}{0.4} = 30\).
Total distance \(= 10 + 30 = 40\) m.
Answer: 40 m.
Example 3 The series \(5 + 5\left(\dfrac{2x-1}{3}\right) + 5\left(\dfrac{2x-1}{3}\right)^2 + \dots\) is geometric. (a) Find the values of \(x\) for which the series converges. (b) Find the sum to infinity when \(x = 1.5\).
(a) The common ratio is \(r = \dfrac{2x-1}{3}\). We need \(-1 \lt \dfrac{2x-1}{3} \lt 1\).
Multiply by 3: \(-3 \lt 2x - 1 \lt 3\). Add 1: \(-2 \lt 2x \lt 4\). Divide by 2: \(-1 \lt x \lt 2\).
(b) When \(x = 1.5\), \(r = \dfrac{3 - 1}{3} = \dfrac{2}{3}\). This is inside the range, so the sum exists.
\(S_\infty = \dfrac{5}{1 - \frac{2}{3}} = \dfrac{5}{\frac{1}{3}} = 15\).
Answer: (a) \(-1 \lt x \lt 2\). (b) 15.
Common errors
- Common error: using \(S_\infty = \dfrac{u_1}{1-r}\) without checking \(|r| \lt 1\). If \(|r| \ge 1\) the series has no sum to infinity.
- Common error: writing the convergence condition as \(r \lt 1\) only. A ratio of \(-3\) also makes the series diverge, so you need \(-1 \lt r \lt 1\).
- Common error: using the wrong first term, such as 10 instead of 12 for the up-and-down distances of a bouncing ball.
- Common error: not simplifying a fraction that comes from a recurring decimal, or leaving \(\dfrac{0.36}{0.99}\) with decimals in it.
- Common error: forgetting the part of a decimal that does not repeat, such as the 2 in \(0.2\overline{7}\).
Exam tip: Always write the check \(|r| \lt 1\) with the actual value of \(r\) before you use the sum to infinity formula. It shows that the formula is valid.
Check yourself
- Find the sum to infinity of \(12 + 6 + 3 + \dots\)
Answer
\(r = \tfrac{1}{2}\), which satisfies \(|r| \lt 1\). \(S_\infty = \dfrac{12}{1 - 0.5} = 24\). - Write \(0.\overline{27}\) as a fraction in its simplest form.
Answer
\(0.27 + 0.0027 + \dots\) has \(u_1 = 0.27\), \(r = 0.01\). \(S_\infty = \dfrac{0.27}{0.99} = \dfrac{27}{99} = \dfrac{3}{11}\). - A convergent geometric series has first term 20 and sum to infinity 50. Find the common ratio.
Answer
\(50 = \dfrac{20}{1 - r}\), so \(1 - r = \dfrac{20}{50} = 0.4\) and \(r = 0.6\). This satisfies \(|r| \lt 1\). - The series \(1 + \dfrac{x}{4} + \dfrac{x^2}{16} + \dots\) is geometric. Find the values of \(x\) for which it converges, and its sum to infinity when \(x = 2\).
Answer
\(r = \dfrac{x}{4}\). Need \(\left|\dfrac{x}{4}\right| \lt 1\), so \(-4 \lt x \lt 4\). For \(x = 2\), \(r = 0.5\) and \(S_\infty = \dfrac{1}{1 - 0.5} = 2\).