1.8 Infinite geometric series

Adding infinitely many positive numbers does not always give an infinite total. The numbers $\frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \ldots$ go on forever, yet their sum never exceeds 1, and adding more and more terms brings the total as close to 1 as desired. Geometric series with a common ratio between $-1$ and $1$ all behave like this. Their sums to infinity describe quantities built from infinitely many shrinking parts, such as the total distance a bouncing ball travels before it comes to rest, and they show that every recurring decimal is a fraction.

From finite sums to infinite sums

A sum that settles down

A square of side 1 has area 1. Shade half of it, then half of what is left, then half of what is still left, and so on. The shaded pieces have areas $\frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16}, \ldots$, each piece half the size of the one before.

After $n$ pieces, the uncovered part has area $\left(\frac{1}{2}\right)^n$, so the shaded area is $1 - \left(\frac{1}{2}\right)^n$. This is never quite 1, but it can be made as close to 1 as desired by taking enough pieces: after 10 pieces the gap is less than $0.001$, and after 20 it is less than one millionth. The infinite sum

is therefore given the value 1.

Partial sums

An infinite sum cannot be calculated by adding all its terms one at a time. Instead, it is studied through its partial sums: the sums of the first term, the first two terms, the first three terms, and so on.

For the series $8 + 4 + 2 + 1 + \cdots$, with $u_1 = 8$ and $r = \frac{1}{2}$, the partial sums are

Each partial sum closes half of the remaining gap to 16. The sequence of partial sums gets as close to 16 as desired, without ever reaching it.

Convergence and divergence

The value that the partial sums approach is called the sum to infinity of the series. The idea of "approaching a value as $n$ increases" is the idea of a limit, which 5.1 develops for functions. Here it is enough to know that $S_n$ gets as close as desired to a fixed number when $n$ is large enough.

A convergent series has a finite sum to infinity even though it has infinitely many terms. A divergent series has no sum to infinity. Three simple examples show the possibilities:

• $1 + \frac{1}{3} + \frac{1}{9} + \cdots$ converges: its partial sums approach $1.5$.

• $1 + 2 + 4 + 8 + \cdots$ diverges: its partial sums $1, 3, 7, 15, \ldots$ grow without bound.

• $1 - 1 + 1 - 1 + \cdots$ diverges: its partial sums $1, 0, 1, 0, \ldots$ jump between two values forever and never settle.

The sum to infinity of a geometric series

Why rⁿ approaches 0 when |r| < 1

The sum of the first $n$ terms of a geometric series, from 1.3, is

The only part of this formula that changes as $n$ grows is $r^n$. So the behaviour of $S_n$ for large $n$ depends entirely on what happens to $r^n$.

When $|r| < 1$, multiplying by $r$ makes a number smaller in size, by the same fraction each time. Repeating this shrinks $r^n$ towards 0, and it eventually falls below any positive number, however small. For example, $0.9^n < 0.001$ as soon as $n > \frac{\ln 0.001}{\ln 0.9} = 65.6$, using the method of 1.7. When $r$ is negative, $r^n$ alternates in sign, but its size $|r|^n$ still shrinks to 0. When $|r| > 1$, the size of $r^n$ grows without bound, and when $r = \pm 1$ it stays at size 1.

Deriving the formula

When $|r| < 1$, the term $r^n$ in $S_n$ approaches 0 as $n$ increases. The numerator $u_1(1 - r^n)$ therefore approaches $u_1(1 - 0) = u_1$, while the denominator $1 - r$ does not change. So the partial sums approach

The formula gives $\frac{8}{1 - 0.5} = 16$ for $8 + 4 + 2 + \cdots$, matching the table of partial sums. The difference between the sum to infinity and a partial sum is

which shows directly how the gap shrinks as $r^n$ does.

What happens when |r| ≥ 1

(In every case $u_1 \neq 0$. If $u_1 = 0$, every term is 0 and the sum is trivially 0.)

Finding sums to infinity

When a series is written in sigma notation, the first term is found by substituting the lower limit, not by reading the number in front of the bracket. In part (d), the 5 is not the first term because the sum starts at $k = 1$, not $k = 0$.

When $r$ is negative the partial sums move alternately above and below the sum to infinity, closing in from both sides. For $8 - 4 + 2 - 1 + \cdots$, with $r = -\frac{1}{2}$, the sum to infinity is $\frac{8}{1.5} = \frac{16}{3}$.

Finding the first term or the common ratio

The formula $S_\infty = \frac{u_1}{1 - r}$ links three quantities, so knowing any two gives the third. Questions often combine it with the formula for a single term, $u_n = u_1 r^{n-1}$, to give simultaneous equations.

How close is a partial sum?

The gap $S_\infty - S_n = \frac{u_1 r^n}{1 - r}$ shows how many terms are needed for a partial sum to be within a given distance of the sum to infinity. This uses the logarithm methods of 1.7.

Series whose common ratio depends on x

Finding where a series converges

When the terms of a geometric series contain a variable, the common ratio may depend on that variable. The series then converges for some values of $x$ and diverges for others. The condition $|r| < 1$ becomes an inequality in $x$, which is best written as a double inequality:

Solving both parts at once gives the interval of values of $x$ for which the sum to infinity exists.

Common ratios that are fractions in x

Sometimes the common ratio has the variable in a denominator, or the terms must first be divided to find $r$. The method is the same, but the inequality needs more care.

Applications of infinite geometric series

Recurring decimals

A recurring decimal is an infinite geometric series in disguise. The decimal $0.777\ldots$ means $0.7 + 0.07 + 0.007 + \cdots$, a series with first term $0.7$ and common ratio $0.1$. A dot over a digit (or over the first and last digits of a repeating block) shows that it repeats: $0.\dot{7} = 0.777\ldots$ and $0.\dot{2}\dot{7} = 0.272727\ldots$

The same method gives a result that often surprises students: $0.\dot{9} = 0.999\ldots = \frac{0.9}{1 - 0.1} = 1$. The decimal $0.999\ldots$ is exactly equal to 1, just as $\frac{2}{2}$ is another way of writing 1. The partial sums $0.9, 0.99, 0.999, \ldots$ are all less than 1, but the infinite sum is their limit, which is exactly 1.

The total distance travelled by a bouncing ball

When a ball is dropped, it bounces back to a fraction of the height it fell from. If that fraction stays the same on every bounce, the heights form a geometric sequence (as in 1.3), and the total distance the ball travels is an infinite geometric series.

The model predicts infinitely many bounces in a finite total distance. In reality a ball stops after a finite number of bounces, because the model ignores effects such as the ball's deformation and air resistance when the bounces become very small. The geometric model is still an accurate approximation for the total distance, since the tiny later bounces contribute almost nothing to it.

Zeno's paradox and infinitely many steps

The Greek philosopher Zeno of Elea argued that motion is impossible: to walk 1 m, a person must first walk half the distance, then half of what remains, then half of that, and so on, completing infinitely many stages. He concluded that the walk could never be finished. The stages have lengths $\frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \ldots$ metres, and their sum to infinity is 1 m. If the person walks at a constant speed and takes $t$ seconds for the first half, the times of the stages are $t, \frac{t}{2}, \frac{t}{4}, \ldots$, with sum $\frac{t}{1 - 1/2} = 2t$ seconds, which is finite. Infinitely many stages can be completed in a finite time because their durations form a convergent series.

The Koch snowflake

A fractal is a shape built by repeating the same construction on smaller and smaller scales. The Koch snowflake starts with an equilateral triangle of side 1. At each stage, the middle third of every side is replaced by two sides of a smaller equilateral triangle pointing outwards.

Each stage replaces every side by 4 sides, each $\frac{1}{3}$ as long. After $n$ stages there are $3 \times 4^n$ sides of length $\left(\frac{1}{3}\right)^n$, so the perimeter is

This is a geometric sequence with $r = \frac{4}{3} > 1$, so the perimeter grows without bound. The area behaves differently, because the area added at each stage forms a convergent geometric series.

The Koch snowflake therefore has a finite area, $\frac{8}{5}$ of the starting triangle, enclosed by a boundary of infinite length.

Exam-style questions on infinite geometric series

Sum to infinity questions appear on both papers. They often begin with finite geometric series from 1.3 and end with a sum to infinity, and the condition $|r| < 1$ is regularly tested, either by rejecting a value of $r$ or by finding the values of $x$ for which a series converges.

Further points that earn or lose marks:

• The formula $S_\infty = \frac{u_1}{1 - r}$ is in the formula booklet; the condition $|r| < 1$ must still be checked or stated.

• A question asking for an exact answer, such as a recurring decimal as a fraction, needs the fraction in its simplest form, not a decimal approximation.

• In contextual questions, such as a bouncing ball, explain how the series was set up (which distances are counted twice) before using the formula.

Self-check

• To do:

  I can explain what a partial sum is and how partial sums decide whether a series converges.

• To do:

  I can explain why $r^n \to 0$ when $|r| < 1$, and derive $S_\infty = \frac{u_1}{1 - r}$ from the formula for $S_n$.

• To do:

  I can decide whether a geometric series converges and find its sum to infinity when it does.

• To do:

  I can find $u_1$ or $r$ from a sum to infinity, rejecting any value of $r$ with $|r| \ge 1$.

• To do:

  I can find the values of $x$ for which a series with a common ratio in terms of $x$ converges.

• To do:

  I can write a recurring decimal as a fraction using a sum to infinity.

• To do:

  I can find the total distance travelled by a bouncing ball, counting each rebound twice.

• To do:

  I can explain how the Koch snowflake has an infinite perimeter but a finite area.