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Geometric series and the sum to infinity

Covers AS91258

The partial sum

The sum of the first $n$ terms of a geometric series is:

$S_n = \frac{a\left(1 - r^{\,n}\right)}{1 - r} \qquad (r \neq 1)$

An equivalent form, more convenient when $r > 1$ because it avoids negatives:

$S_n = \frac{a\left(r^{\,n} - 1\right)}{r - 1}$

Each part:
- $a$ — the first term
- $r$ — the common ratio
- $n$ — the number of terms being added
The exponent on $r$ here is $n$ , not $n-1$ . That differs from the general term, and swapping them is a frequent error.
The two forms give identical answers — multiplying top and bottom of one by $-1$ gives the other. Use whichever avoids negative numbers.

Working through an example

Find the sum of the first 8 terms of $2, 6, 18, \ldots$ $2, 6, 18, \dots$
- $a = 2$ , $r = 3$ , $n = 8$
- Since $r > 1$ , use the second form:

$S_8 = \frac{2\left(3^8 - 1\right)}{3 - 1} = \frac{2(6561 - 1)}{2} = 6560$

Check with the other form: $\dfrac{2\left(1 - 6561\right)}{1-3} = \dfrac{2(-6560)}{-2} = 6560$ ✓

The sum to infinity

Adding infinitely many terms usually gives an infinite total — but not always.
If the terms shrink fast enough, the running total settles on a finite value. The series is then said to converge.

$S_\infty = \frac{a}{1 - r} \qquad \text{provided } |r| < 1$

The condition $|r| < 1$ is essential. It means $-1 < r < 1$ : the ratio must be strictly between $-1$ and $1$ .
State the condition and check it before using the formula. Applying $S_\infty$ when $|r| \geq 1$ produces a meaningless number.

Why the condition works

Look at the partial sum formula: $S_n = \dfrac{a\left(1 - r^n\right)}{1-r}$ .
If $|r| < 1$ , then $r^n \to 0$ as $n$ grows — repeatedly multiplying by a fraction drives the term towards nothing. The formula tends to $\dfrac{a(1-0)}{1-r} = \dfrac{a}{1-r}$ .
If $|r| \geq 1$ , then $r^n$ does not settle — it grows without bound (or oscillates), so the sum has no finite limit.
Worked through — $8 + 4 + 2 + 1 + \ldots$
- $a = 8$ and $r = \frac{1}{2}$ , and $\left|\frac{1}{2}\right| < 1$ ✓ so the series converges
- $S_\infty = \dfrac{8}{1 - 0.5} = \dfrac{8}{0.5} = 16$
- The running totals are $8, 12, 14, 15, 15.5, 15.75, \ldots$ — visibly closing in on 16 without reaching it.

Working backwards from $S_\infty$

Questions often give the sum to infinity and ask for $a$ or $r$ .
Substitute into $S_\infty = \dfrac{a}{1-r}$ and solve.
If two pieces of information are given, you may get a quadratic — and then both roots must be tested against $|r| < 1$ , rejecting any that fails.

Contexts for the sum to infinity

Total distance travelled by a bouncing ball before it stops
Total length of an infinitely repeating pattern
Recurring decimals — $0.\dot{3}$ is $0.3 + 0.03 + 0.003 + \cdots$ , a geometric series with $a = 0.3$ and $r = 0.1$ , summing to $\dfrac{0.3}{0.9} = \dfrac{1}{3}$
The bouncing ball needs care. After the drop, the ball travels each rebound height twice — once up and once down — so the total distance is the drop plus twice the sum of the rebound heights.

Tips and Tricks

Check $|r| < 1$ before using $S_\infty$ , and say that you have. If $|r| \geq 1$ the series diverges and has no sum to infinity — that statement is the answer, not a failure.

The exponent in the sum formula is $n$ , not $n-1$ . The general term uses $r^{n-1}$ ; the sum uses $r^n$ . Keeping the two straight is worth a deliberate check.

Choose the form that avoids negatives. For $r > 1$ use $\frac{a(r^n - 1)}{r-1}$ ; for $r < 1$ use $\frac{a(1-r^n)}{1-r}$ . Both are correct, but one has fewer sign errors.

For a bouncing ball, count the up-and-down. Total distance = drop height + $2 \times$ (sum of rebound heights). Forgetting the factor of 2 is the standard error in these questions.

If a quadratic gives two values of $r$ , test both against $|r| < 1$ and reject with a reason. A ratio of 1.5 cannot produce a finite sum to infinity.

A recurring decimal is a geometric series. Recognising that turns "convert $0.\dot{4}\dot{5}$ to a fraction" into a one-line application of $S_\infty$ .

Worked ExampleA bouncing ball and its total distance

A ball is dropped from a height of 2 m and rebounds to 60% of its previous height after each bounce. Find the total vertical distance travelled by the ball before it comes to rest.

Step 1 — Picture the journey carefully

The ball's path is:

Down 2 m (the initial drop)
Up 1.2 m, then down 1.2 m (first rebound)
Up 0.72 m, then down 0.72 m (second rebound)
and so on

Step 2 — Separate the two parts

$\text{Total distance} = \underbrace{2}_{\text{initial drop}} + \underbrace{2 \times \left(\text{sum of all rebound heights}\right)}_{\text{up and down}}$

Step 3 — Identify the rebound heights as a geometric series

The rebound heights are:

$1.2, \; 0.72, \; 0.432, \; \ldots$

Test the ratio:

$\frac{0.72}{1.2} = 0.6 \qquad \frac{0.432}{0.72} = 0.6$

Constant, so it is geometric with

$a = 2 \times 0.6 = 1.2, \qquad r = 0.6$

Step 4 — Check the convergence condition

$|r| = 0.6 < 1 \;\checkmark$

The series converges, so the sum to infinity exists and the total distance is finite — the ball travels a limited distance despite bouncing infinitely often.

Step 5 — Find the sum of the rebound heights

$S_\infty = \frac{a}{1-r} = \frac{1.2}{1 - 0.6} = \frac{1.2}{0.4} = 3$

The rebound heights total 3 m.

Step 6 — Assemble the total distance

$\text{Total} = 2 + 2(3) = 2 + 6 = 8$

Step 7 — Sanity-check by partial sums

Adding the first few stages directly:

Drop: 2 m — running total 2
Bounce 1: $2 \times 1.2 = 2.4$ — total 4.4
Bounce 2: $2 \times 0.72 = 1.44$ — total 5.84
Bounce 3: $2 \times 0.432 = 0.864$ — total 6.70
Bounce 4: $2 \times 0.2592 = 0.518$ — total 7.22
Bounce 5: $0.311$ — total 7.53

The running total is climbing towards 8 and slowing down ✓ consistent with the answer.

Step 8 — Interpret the result

Although the ball bounces infinitely many times in the model, the total distance is finite at 8 m, because the bounce heights shrink geometrically.

The model's limitation: the ball also takes a finite total time, but in reality it stops bouncing after a limited number of bounces once the rebound energy falls below what the surface can return. The infinite model is a good approximation, not a literal description.

Geometric series and the sum to infinity

Covers AS91258

The partial sum

The sum of the first $n$ terms of a geometric series is:

$S_n = \frac{a\left(1 - r^{\,n}\right)}{1 - r} \qquad (r \neq 1)$

An equivalent form, more convenient when $r > 1$ because it avoids negatives:

$S_n = \frac{a\left(r^{\,n} - 1\right)}{r - 1}$

Each part:
- $a$ — the first term
- $r$ — the common ratio
- $n$ — the number of terms being added
The exponent on $r$ here is $n$ , not $n-1$ . That differs from the general term, and swapping them is a frequent error.
The two forms give identical answers — multiplying top and bottom of one by $-1$ gives the other. Use whichever avoids negative numbers.

Working through an example

Find the sum of the first 8 terms of $2, 6, 18, \ldots$ $2, 6, 18, \dots$
- $a = 2$ , $r = 3$ , $n = 8$
- Since $r > 1$ , use the second form:

$S_8 = \frac{2\left(3^8 - 1\right)}{3 - 1} = \frac{2(6561 - 1)}{2} = 6560$

Check with the other form: $\dfrac{2\left(1 - 6561\right)}{1-3} = \dfrac{2(-6560)}{-2} = 6560$ ✓

The sum to infinity

Adding infinitely many terms usually gives an infinite total — but not always.
If the terms shrink fast enough, the running total settles on a finite value. The series is then said to converge.

$S_\infty = \frac{a}{1 - r} \qquad \text{provided } |r| < 1$

The condition $|r| < 1$ is essential. It means $-1 < r < 1$ : the ratio must be strictly between $-1$ and $1$ .
State the condition and check it before using the formula. Applying $S_\infty$ when $|r| \geq 1$ produces a meaningless number.

Why the condition works

Look at the partial sum formula: $S_n = \dfrac{a\left(1 - r^n\right)}{1-r}$ .
If $|r| < 1$ , then $r^n \to 0$ as $n$ grows — repeatedly multiplying by a fraction drives the term towards nothing. The formula tends to $\dfrac{a(1-0)}{1-r} = \dfrac{a}{1-r}$ .
If $|r| \geq 1$ , then $r^n$ does not settle — it grows without bound (or oscillates), so the sum has no finite limit.
Worked through — $8 + 4 + 2 + 1 + \ldots$
- $a = 8$ and $r = \frac{1}{2}$ , and $\left|\frac{1}{2}\right| < 1$ ✓ so the series converges
- $S_\infty = \dfrac{8}{1 - 0.5} = \dfrac{8}{0.5} = 16$
- The running totals are $8, 12, 14, 15, 15.5, 15.75, \ldots$ — visibly closing in on 16 without reaching it.

Working backwards from $S_\infty$

Questions often give the sum to infinity and ask for $a$ or $r$ .
Substitute into $S_\infty = \dfrac{a}{1-r}$ and solve.
If two pieces of information are given, you may get a quadratic — and then both roots must be tested against $|r| < 1$ , rejecting any that fails.

Contexts for the sum to infinity

Total distance travelled by a bouncing ball before it stops
Total length of an infinitely repeating pattern
Recurring decimals — $0.\dot{3}$ is $0.3 + 0.03 + 0.003 + \cdots$ , a geometric series with $a = 0.3$ and $r = 0.1$ , summing to $\dfrac{0.3}{0.9} = \dfrac{1}{3}$
The bouncing ball needs care. After the drop, the ball travels each rebound height twice — once up and once down — so the total distance is the drop plus twice the sum of the rebound heights.

Tips and Tricks

Check $|r| < 1$ before using $S_\infty$ , and say that you have. If $|r| \geq 1$ the series diverges and has no sum to infinity — that statement is the answer, not a failure.

The exponent in the sum formula is $n$ , not $n-1$ . The general term uses $r^{n-1}$ ; the sum uses $r^n$ . Keeping the two straight is worth a deliberate check.

Choose the form that avoids negatives. For $r > 1$ use $\frac{a(r^n - 1)}{r-1}$ ; for $r < 1$ use $\frac{a(1-r^n)}{1-r}$ . Both are correct, but one has fewer sign errors.

For a bouncing ball, count the up-and-down. Total distance = drop height + $2 \times$ (sum of rebound heights). Forgetting the factor of 2 is the standard error in these questions.

If a quadratic gives two values of $r$ , test both against $|r| < 1$ and reject with a reason. A ratio of 1.5 cannot produce a finite sum to infinity.

A recurring decimal is a geometric series. Recognising that turns "convert $0.\dot{4}\dot{5}$ to a fraction" into a one-line application of $S_\infty$ .

Worked ExampleA bouncing ball and its total distance

A ball is dropped from a height of 2 m and rebounds to 60% of its previous height after each bounce. Find the total vertical distance travelled by the ball before it comes to rest.

Step 1 — Picture the journey carefully

The ball's path is:

Down 2 m (the initial drop)
Up 1.2 m, then down 1.2 m (first rebound)
Up 0.72 m, then down 0.72 m (second rebound)
and so on

Step 2 — Separate the two parts

$\text{Total distance} = \underbrace{2}_{\text{initial drop}} + \underbrace{2 \times \left(\text{sum of all rebound heights}\right)}_{\text{up and down}}$

Step 3 — Identify the rebound heights as a geometric series

The rebound heights are:

$1.2, \; 0.72, \; 0.432, \; \ldots$

Test the ratio:

$\frac{0.72}{1.2} = 0.6 \qquad \frac{0.432}{0.72} = 0.6$

Constant, so it is geometric with

$a = 2 \times 0.6 = 1.2, \qquad r = 0.6$

Step 4 — Check the convergence condition

$|r| = 0.6 < 1 \;\checkmark$

The series converges, so the sum to infinity exists and the total distance is finite — the ball travels a limited distance despite bouncing infinitely often.

Step 5 — Find the sum of the rebound heights

$S_\infty = \frac{a}{1-r} = \frac{1.2}{1 - 0.6} = \frac{1.2}{0.4} = 3$

The rebound heights total 3 m.

Step 6 — Assemble the total distance

$\text{Total} = 2 + 2(3) = 2 + 6 = 8$

Step 7 — Sanity-check by partial sums

Adding the first few stages directly:

Drop: 2 m — running total 2
Bounce 1: $2 \times 1.2 = 2.4$ — total 4.4
Bounce 2: $2 \times 0.72 = 1.44$ — total 5.84
Bounce 3: $2 \times 0.432 = 0.864$ — total 6.70
Bounce 4: $2 \times 0.2592 = 0.518$ — total 7.22
Bounce 5: $0.311$ — total 7.53

The running total is climbing towards 8 and slowing down ✓ consistent with the answer.

Step 8 — Interpret the result

Although the ball bounces infinitely many times in the model, the total distance is finite at 8 m, because the bounce heights shrink geometrically.

Geometric series and the sum to infinity

The partial sum

Working through an example

The sum to infinity

Why the condition works

Working backwards from S∞S_\inftyS∞​

Contexts for the sum to infinity

Step 1 — Picture the journey carefully

Step 2 — Separate the two parts

Step 3 — Identify the rebound heights as a geometric series

Step 4 — Check the convergence condition

Step 5 — Find the sum of the rebound heights

Step 6 — Assemble the total distance

Step 7 — Sanity-check by partial sums

Step 8 — Interpret the result

Geometric series and the sum to infinity

The partial sum

Working through an example

The sum to infinity

Why the condition works

Working backwards from S∞S_\inftyS∞​

Contexts for the sum to infinity

Step 1 — Picture the journey carefully

Step 2 — Separate the two parts

Step 3 — Identify the rebound heights as a geometric series

Step 4 — Check the convergence condition

Step 5 — Find the sum of the rebound heights

Step 6 — Assemble the total distance

Step 7 — Sanity-check by partial sums

Step 8 — Interpret the result

Working backwards from $S_\infty$

Working backwards from $S_\infty$