2.4 Sum to Infinity
SL 1.8
- what happens to \(S_n\) of a geometric series as \(n\) becomes very large
- when a geometric series is convergent, and the condition \(|r| < 1\)
- the formula for the sum to infinity, \(S_\infty = \dfrac{u_1}{1 - r}\)
- how to find the values of \(x\) for which a series in \(x\) converges
- how to write a recurring decimal as a fraction
What happens as \(n\) grows?
Take the geometric series with first term \(u_1 = 1\). Compare two common ratios.
| \(n\) | 1 | 2 | 3 | 5 | 10 | 20 |
|---|---|---|---|---|---|---|
| \(S_n\) when \(r = 2\) | 1 | 3 | 7 | 31 | 1023 | 1 048 575 |
| \(S_n\) when \(r = 0.5\) | 1 | 1.5 | 1.75 | 1.9375 | 1.998… | 1.999 998… |
- When \(r = 2\), the sum grows without limit. We say the series diverges.
- When \(r = 0.5\), the sum gets closer and closer to \(2\), but never passes it. We say the series converges, and its sum to infinity is \(2\).
Why? Look at the formula \(S_n = \dfrac{u_1 (1 - r^n)}{1 - r}\). If \(-1 < r < 1\), then \(r^n\) gets closer and closer to \(0\) as \(n\) grows (for example \(0.5^{20} \approx 0.000\,001\)). So \(S_n\) gets closer to \(\dfrac{u_1(1 - 0)}{1 - r}\).
Lists & Spreadsheet page: in column A generate \(n = 1, 2, \ldots, 30\) (Menu → Data → Generate Sequence, u(n)=u(n-1)+1, initial term 1). In the formula cell of column B type =(1-0.5^a[])/(1-0.5). Scroll down: the values approach \(2\). Change \(0.5\) to \(2\) and watch them grow.
The sum to infinity
Sum to infinity of a geometric series In the formula booklet
\[ S_\infty = \frac{u_1}{1 - r}, \quad |r| < 1 \]
A geometric series converges if and only if \(|r| < 1\), that is, \(-1 < r < 1\).
- \(S_\infty\): the sum to infinity
- \(u_1\): the first term
- \(r\): the common ratio
If \(|r| \ge 1\), the sum to infinity does not exist. Always check the condition first.
Example 1 Find the sum to infinity of the geometric series \(24 - 8 + \dfrac{8}{3} - \cdots\)
\(u_1 = 24\) and \(r = \dfrac{-8}{24} = -\dfrac{1}{3}\). Since \(\left|-\frac{1}{3}\right| < 1\), the series converges.
\[ S_\infty = \frac{24}{1 - \left(-\frac{1}{3}\right)} = \frac{24}{\frac{4}{3}} = 18 \]
Series that contain \(x\)
When the ratio contains \(x\), the condition \(|r| < 1\) becomes an inequality in \(x\).
Example 2 Consider the geometric series \(4 - 2x + x^2 - \dfrac{x^3}{2} + \cdots\)
(a) Find the range of values of \(x\) for which the series converges.
(b) Find the sum to infinity in terms of \(x\).
(a) \(r = \dfrac{-2x}{4} = -\dfrac{x}{2}\). The series converges when
\[ \left|-\frac{x}{2}\right| < 1 \;\Longrightarrow\; |x| < 2 \;\Longrightarrow\; -2 < x < 2 . \]
(b)
\[ S_\infty = \frac{4}{1 - \left(-\frac{x}{2}\right)} = \frac{4}{1 + \frac{x}{2}} = \frac{8}{2 + x}, \qquad -2 < x < 2 . \]
Using \(S_n\) and \(S_\infty\) together
Example 3 The sum of the first three terms of a geometric series is \(234\), and the sum to infinity is \(250\).
(a) Find the common ratio, and show that the first term is \(150\).
(b) Find the least value of \(n\) for which \(S_\infty - S_n < 0.01\).
(a) Divide \(S_3\) by \(S_\infty\). Most of the formula cancels:
\[ \frac{S_3}{S_\infty} = \frac{\dfrac{u_1(1 - r^3)}{1 - r}}{\dfrac{u_1}{1 - r}} = 1 - r^3 = \frac{234}{250} = 0.936 . \]
So \(r^3 = 0.064\), giving \(r = 0.4\). Then \(u_1 = S_\infty(1 - r) = 250 \times 0.6 = 150\).
(b)
\[ S_\infty - S_n = \frac{u_1}{1 - r} - \frac{u_1(1 - r^n)}{1 - r} = \frac{u_1 r^n}{1 - r} = 250 \times 0.4^n . \]
Solve \(250 \times 0.4^n < 0.01\), so \(0.4^n < 0.000\,04\). With logarithms (or a GDC table):
\[ n > \frac{\ln 0.000\,04}{\ln 0.4} = 11.05\ldots \]
Dividing by \(\ln 0.4 < 0\) reverses the inequality. The least value is \(n = 12\).
Recurring decimals
A recurring decimal is the sum of an infinite geometric series, so we can write it as a fraction.
Example 4 Use an infinite geometric series to show that \(0.2\dot{3}\dot{6} = 0.236\,363\,6\ldots = \dfrac{13}{55}\).
Separate the part that does not repeat:
\[ 0.236\,363\,6\ldots = 0.2 + \underbrace{0.036 + 0.000\,36 + 0.000\,003\,6 + \cdots}_{\text{geometric}} \]
The geometric part has \(u_1 = 0.036\) and \(r = 0.01\), with \(|r| < 1\):
\[ \frac{0.036}{1 - 0.01} = \frac{0.036}{0.99} = \frac{36}{990} = \frac{2}{55} . \]
So \(0.2\dot{3}\dot{6} = \dfrac{1}{5} + \dfrac{2}{55} = \dfrac{11}{55} + \dfrac{2}{55} = \dfrac{13}{55}\).
Common errors
| ✗ Wrong | ✓ Right | Why |
|---|---|---|
| \(S_\infty = \dfrac{u_1}{1 - r}\) for \(r = 3\) | \(S_\infty\) does not exist | The formula needs \(\lvert r \rvert < 1\). |
| \(\dfrac{24}{1 - \frac{1}{3}}\) for \(r = -\frac{1}{3}\) | \(\dfrac{24}{1 + \frac{1}{3}}\) | Substitute the negative ratio with its sign. |
| \(\lvert x \rvert < 2\) written as \(x < 2\) | \(-2 < x < 2\) | The absolute value gives two inequalities. |
| \(0.4^n < 0.00004 \Rightarrow n < 11.05\) | \(n > 11.05\) | Dividing by \(\ln 0.4\) (negative) reverses the sign. |
| An arithmetic series has a sum to infinity | Only geometric series with \(\lvert r \rvert < 1\) do | An arithmetic series with \(d \neq 0\) always grows without limit. |
Exercises
1 Find the sum to infinity of \(18 + 6 + 2 + \cdots\)
\(r = \dfrac{1}{3}\). \(\;S_\infty = \dfrac{18}{1 - \frac{1}{3}} = \dfrac{18}{\frac{2}{3}} = 27\).
2 Find the sum to infinity of \(40 - 30 + 22.5 - \cdots\)
\(r = -\dfrac{30}{40} = -0.75\). \(\;S_\infty = \dfrac{40}{1.75} = \dfrac{160}{7}\).
3 Explain why the geometric series \(2 + 3 + 4.5 + \cdots\) has no sum to infinity.
\(r = \dfrac{3}{2} = 1.5\), and \(|1.5| \ge 1\). The terms get bigger, so the sum grows without limit.
4 A geometric series has sum to infinity \(60\) and first term \(15\). Find the common ratio.
\(\dfrac{15}{1 - r} = 60 \Rightarrow 1 - r = \dfrac{1}{4} \Rightarrow r = \dfrac{3}{4}\).
5 A geometric series has common ratio \(-\dfrac{2}{3}\) and sum to infinity \(36\). Find the first term and the second term.
\(u_1 = S_\infty(1 - r) = 36 \times \dfrac{5}{3} = 60\), and \(u_2 = 60 \times \left(-\dfrac{2}{3}\right) = -40\).
6 For the geometric series \(1 + 3x + 9x^2 + \cdots\)
(a) find the values of \(x\) for which it converges,
(b) find \(x\) when the sum to infinity is \(4\).
(a) \(r = 3x\), so \(|3x| < 1\), that is \(-\dfrac{1}{3} < x < \dfrac{1}{3}\).
(b) \(\dfrac{1}{1 - 3x} = 4 \Rightarrow 1 - 3x = \dfrac{1}{4} \Rightarrow x = \dfrac{1}{4}\), which is inside the range ✓.
7 For the geometric series \(5 + 5(x - 1) + 5(x - 1)^2 + \cdots\), find the values of \(x\) for which it converges, and the sum to infinity in terms of \(x\).
\(r = x - 1\), so \(|x - 1| < 1\), that is \(0 < x < 2\).
\(S_\infty = \dfrac{5}{1 - (x - 1)} = \dfrac{5}{2 - x}\).
8 Write \(0.\dot{4}\dot{5} = 0.454\,545\ldots\) as a fraction in its lowest terms.
\(0.45 + 0.0045 + \cdots\) has \(u_1 = 0.45\), \(r = 0.01\): \(\;\dfrac{0.45}{0.99} = \dfrac{45}{99} = \dfrac{5}{11}\).
9 A ball is dropped from a height of 8 m. After each bounce it rises to 75% of its previous height. Find the total vertical distance the ball travels before it comes to rest.
Down: \(8\) m first. After that, each bounce goes up and comes down the same height:
\[ 8 + 2(6 + 4.5 + 3.375 + \cdots) = 8 + 2 \times \frac{6}{1 - 0.75} = 8 + 2 \times 24 = 56 \text{ m}. \]
10 The first term of a geometric series is \(12\) and its sum to infinity is \(16\). Find the least value of \(n\) for which \(S_n > 15.99\).
\(1 - r = \dfrac{12}{16}\), so \(r = \dfrac{1}{4}\). Then \(S_n = 16\left(1 - \left(\frac{1}{4}\right)^n\right)\).
\(16\left(1 - 0.25^n\right) > 15.99 \Rightarrow 0.25^n < \dfrac{0.01}{16} = 0.000\,625\).
\(0.25^5 \approx 0.000\,977\) and \(0.25^6 \approx 0.000\,244\), so the least value is \(n = 6\).