1.8 Sum of infinite geometric sequences · Educerie Mathematics: analysis and approaches

Educerie · IB Diploma · Mathematics: analysis and approaches

Topic 1 Number and algebra · 1.8 Sum of infinite geometric sequences

Level: SL and HL. Nothing here is HL only, so every section is examinable for both.
Themes (key concepts): patterns, generalization, quantity. The pattern of a geometric series, each term a fixed fraction of the last, lets you generalize from the sum of n terms to the sum of infinitely many, and find a finite quantity where you might expect none.
The question this unit answers: how can you add up infinitely many numbers and get a finite answer, and when can you not?
Where it is examined: Paper 1 (no calculator), usually as a short question of 4 to 6 marks: find S∞, find r or u₁ from S∞, or find the values of x for which a series converges. Paper 2, inside longer sequences questions that compare S∞ with Sₙ or model a real process such as a bouncing ball or a repeated dose. HL Paper 3 uses it as a tool in longer problems.

What you must be able to do

You must be able to	Level	What it looks like in the exam
State the condition for a geometric series to converge, using modulus notation	SL, HL	"Explain why the sum to infinity exists" (1 mark): because −1 < r < 1
Find S∞ = u₁ ÷ (1 − r) from given terms	SL, HL	"Find the sum to infinity of 54 + 18 + 6 + …" (3 marks)
Work backwards from S∞ to r or u₁, including through a quadratic	SL, HL	"The sum to infinity is 32 and the second term is 6. Find the possible values of r" (5 to 6 marks)
Find the values of x for which a series in x converges, and its sum	SL, HL	"Find the values of x for which the sum to infinity exists" (3 marks), then "find x given S∞ = 5"
Compare S∞ with Sₙ and find how many terms are needed	SL, HL	Paper 2: "Find the least n for which S∞ − Sₙ < 0.01" (4 marks)
Use S∞ in context: recurring decimals, bouncing balls, repeated processes	SL, HL	"Find the total distance travelled by the ball" (4 marks)

Before you start

You need geometric sequences and series from 1.3: the common ratio r = u₂ ÷ u₁, the nth term uₙ = u₁rⁿ⁻¹, and the sum of the first n terms, Sₙ = u₁(1 − rⁿ) ÷ (1 − r), which is in the formula booklet. For section 5 you also need to solve an equation with n in the exponent using logarithms (1.7), and for section 6, a linear inequality.

1The idea in one paragraph

Add 8 + 4 + 2 + 1 + ½ + … for ever. Each term is half the one before, so each one covers half of whatever gap is left below 16. The running total creeps up to 16 and never passes it, and "the sum to infinity" means the number the running totals close in on. That only happens when the terms shrink fast enough, which for a geometric series means the common ratio r lies strictly between −1 and 1, written |r| < 1. When it does, the sum is S∞ = u₁ ÷ (1 − r), which is in the formula booklet. When it does not, the series has no sum to infinity at all. The whole subtopic is using that formula, and knowing when you are allowed to.

2Where the formula comes from

Start from the sum of the first n terms, from 1.3:

S_n = u₁(1 − rⁿ) / (1 − r)

The only part that depends on n is rⁿ. Watch what it does as n grows.

n	1	2	5	10	20
(0.5)ⁿ	0.5	0.25	0.03125	0.000977	0.00000095
(−0.5)ⁿ	−0.5	0.25	−0.03125	0.000977	0.00000095
(1.2)ⁿ	1.2	1.44	2.49	6.19	38.3

When |r| < 1, rⁿ shrinks towards 0 (the negative one flips sign each time, but its size still shrinks). So the bracket (1 − rⁿ) tends to 1, and Sₙ tends to u₁ ÷ (1 − r). When |r| > 1, rⁿ grows without limit and so does Sₙ.

Figure 1 plots the running totals, or partial sums, for 8 + 4 + 2 + 1 + …, where u₁ = 8 and r = 0.5. They go 8, 12, 14, 15, 15.5, and each one halves the distance still left to 16.

Figure 1 · The partial sums of 8 + 4 + 2 + 1 + … close in on 16

The formula agrees: S∞ = 8 ÷ (1 − 0.5) = 8 ÷ 0.5 = 16.

A series whose partial sums close in on a single number is said to converge, and that number is its sum to infinity. A series whose partial sums do not settle on a single number diverges, and has no sum to infinity.

Figure 2 shows the same idea with areas. Take a square of area 1. Shade half of it, then half of what is left, then half of what is left again. The pieces are ½, ¼, ⅛, …, a geometric series with u₁ = ½ and r = ½. Every piece fits inside the square, and between them they fill it, so the sum is 1. The formula says the same: ½ ÷ (1 − ½) = 1.

Figure 2 · 1/2 + 1/4 + 1/8 + … fills one square and no more

3The formula and its condition

S∞ = u₁ ÷ (1 − r), and only when |r| < 1.

|r| is the modulus of r, its size with the sign ignored: |−0.5| = 0.5 and |0.5| = 0.5. So |r| < 1 is the short way to write −1 < r < 1. Figure 3 draws it on a number line. Notice that the end points are open circles: r = 1 and r = −1 are both excluded.

Figure 3 · The sum to infinity exists only when |r| < 1

Why the ends fail:

r = 1: 5 + 5 + 5 + … grows without limit. (The formula would divide by zero, which is the formula's own warning.)
r = −1: 5 − 5 + 5 − 5 + … has partial sums 5, 0, 5, 0, … that jump for ever and never settle. It does not converge, even though it does not grow.

Worked example 1. Paper 1. Find the sum to infinity of 10 + 7.5 + 5.625 + ….

r = 7.5 ÷ 10 = 0.75always find r from the terms first

|r| = 0.75 < 1, so the sum to infinity exists

S∞ = 10 ÷ (1 − 0.75) = 10 ÷ 0.25 = 40

The middle line is not decoration. A question that says "find the sum to infinity" assumes it exists, but a question that says "explain why the sum to infinity exists" wants exactly that line, and it is worth a mark.

4Negative ratios, and ratios that are too big

A negative ratio makes the terms alternate in sign. If |r| < 1 the series still converges; the partial sums just approach the limit from alternate sides.

Worked example 2. Find the sum to infinity of 12 − 6 + 3 − 1.5 + ….

r = −6 ÷ 12 = −0.5, and |−0.5| < 1

S∞ = 12 ÷ (1 − (−0.5)) = 12 ÷ 1.5 = 8

The bracket is where this one goes wrong: 1 − (−0.5) is 1.5, not 0.5. Figure 4(a) shows the partial sums, 12, 6, 9, 7.5, 8.25, 7.875, …, landing above and below 8 alternately and closer each time. Figure 4(b) is a series with r = 1.2. Its terms grow, so its partial sums run away and there is nothing to close in on.

Figure 4 · A negative ratio still converges; a ratio above 1 never does

If you ever find S∞ coming out negative for a series whose terms are all positive, or smaller than the first term for a series with positive r, stop. You have almost certainly used an r with |r| ≥ 1, where the formula does not apply, or lost a sign in the bracket.

5Using the formula in both directions

The formula links three quantities, u₁, r and S∞. Give any two and the third follows.

Worked example 3. Paper 1. A geometric series has second term 6 and sum to infinity 32. Find the possible values of r, and the first term for each.

u₁ r = 6second term

u₁ / (1 − r) = 32 → u₁ = 32(1 − r)sum to infinity

32(1 − r) r = 6substitute for u₁

32r − 32r² = 6

16r² − 16r + 3 = 0rearrange and divide by 2

(4r − 1)(4r − 3) = 0

r = 1/4 or r = 3/4

Both satisfy |r| < 1, so both are valid.

r = 1/4: u₁ = 32 × 3/4 = 24; r = 3/4: u₁ = 32 × 1/4 = 8

Two series fit: 24 + 6 + 1.5 + … and 8 + 6 + 4.5 + …. Check both: 24 ÷ (3/4) = 32 and 8 ÷ (1/4) = 32. When a quadratic gives you values of r, test each one against |r| < 1. Often one of them fails and must be rejected, with the reason written down.

How close is Sₙ to S∞? Subtract the two formulas and most of it cancels:

S∞ − S_n = u₁/(1 − r) − u₁(1 − rⁿ)/(1 − r)

= u₁ rⁿ / (1 − r)

That is the gap still left after n terms, and it is how Paper 2 asks "how many terms are needed".

Worked example 4. Paper 2. For 8 + 4 + 2 + …, find the least number of terms for which S∞ − Sₙ < 0.01.

S∞ − S_n = 8 × 0.5ⁿ / 0.5 = 16 × 0.5ⁿ

16 × 0.5ⁿ < 0.01

0.5ⁿ < 0.000625

n ln 0.5 < ln 0.000625

n > ln 0.000625 ÷ ln 0.5 = 10.6dividing by ln 0.5 < 0 reverses the inequality

n = 11

Check: n = 10 gives a gap of 0.0156; n = 11 gives 0.0078 < 0.01

On Paper 2 you can instead tabulate 16 × 0.5ⁿ on the GDC and read off the first n below 0.01. Either way, give a whole number, and check the value either side of it.

6When r contains x

Some series have x in the ratio, and the question asks which values of x make the series converge. The method is always the same: find r in terms of x, write |r| < 1, and solve.

Worked example 5. Paper 1. Consider the series 1 + (x − 1)/2 + ((x − 1)/2)² + …. (a) Find the values of x for which the sum to infinity exists. (b) Find the sum to infinity in terms of x. (c) Find x when the sum to infinity is 4.

(a) r = (x − 1)/2

|(x − 1)/2| < 1

−1 < (x − 1)/2 < 1

−2 < x − 1 < 2

−1 < x < 3

(b) S∞ = 1 / (1 − (x − 1)/2) = 2 / (2 − (x − 1)) = 2 / (3 − x)

3 − x = 1/2

x = 5/2, which lies in −1 < x < 3, so it is valid

Figure 5 shows how the interval for r turns into the interval for x. The answer to (a) is an interval, written with two strict inequalities, not a single value and not "x < 3".

Figure 5 · Turning |r| < 1 into an interval for x

Part (c) ends with a check against part (a). If an answer to (c) fell outside the interval, the series with that x would not converge, and the answer would have to be rejected.

A shorter case: for 3 + 6x + 12x² + …, r = 2x, so |2x| < 1 and |x| < ½, that is, −½ < x < ½.

7Sums to infinity in context

Recurring decimals. Every recurring decimal is a geometric series in disguise, which is how you prove it is a fraction.

0.454545… = 0.45 + 0.0045 + 0.000045 + …

u₁ = 0.45, r = 0.01each block is one hundredth of the one before

S∞ = 0.45 / (1 − 0.01) = 0.45 / 0.99 = 45/99 = 5/11

A bouncing ball. A ball is dropped from 3 m and each time rises to 60% of the height it fell from. How far does it travel in total, in theory, before coming to rest? Figure 6 draws it.

Figure 6 · A ball dropped from 3 m, rising to 60% of each height

The first 3 m is travelled once, downwards. Every later height is travelled twice, up and then down again. The rebound heights are 1.8, 1.08, 0.648, …, a geometric series with u₁ = 1.8 and r = 0.6.

Total = 3 + 2 × (1.8 + 1.08 + 0.648 + …)

= 3 + 2 × 1.8 / (1 − 0.6)

= 3 + 2 × 4.5

= 12 m

The trap is the first drop. Students either double it (as if the ball had been thrown up to 3 m first) or put the 3 inside the series. Draw the picture, and the structure is obvious.

A thought about infinity. Nobody can add up infinitely many numbers, one by one, in finite time, and nobody has ever watched a ball bounce infinitely often. Yet the sum is exact. What mathematics actually establishes is a statement about finite sums: they get as close to 16 as you like, and never exceed it. "The sum to infinity" is a name for that limit. Whether that counts as knowing something about infinity itself, or only about ever-longer finite lists, is a fair question for TOK.

8Where marks are lost

Using S∞ when |r| ≥ 1. The formula gives a number for r = 2 (u₁ ÷ (1 − 2) = −u₁), and the number is meaningless. Check |r| < 1 every time, and write it.

Losing the sign in 1 − r. With r = −0.5, 1 − r is 1.5, not 0.5.

Including the end points. |r| < 1 is strict. r = 1 and r = −1 both diverge, so an interval for x must use < and not ≤.

Writing only half of the interval. |(x − 1)/2| < 1 gives −1 < x < 3, not just x < 3. The modulus opens into two inequalities.

Not checking a found value against the interval. A value of x or r found from S∞ must be tested against |r| < 1, and rejected, with a reason, if it fails.

Mixing up Sₙ and S∞. "The sum of the first 10 terms" is S₁₀ from 1.3; "the sum to infinity" is S∞. Read which one the question wants.

Doubling the first drop. In a bouncing problem, the first fall happens once; every later height happens twice.

9Work it right

Find r first, by dividing a term by the one before it. Write it.
Write |r| < 1 with the value of r, so the marker sees you checked.
Substitute into S∞ = u₁ ÷ (1 − r), with a bracket round a negative r.
For "which values of x", write |r| < 1, open it into −1 < r < 1, and solve for x as a double inequality. Use strict signs.
Working backwards to r through a quadratic, test every root against |r| < 1 and reject any that fail, in words.
For "how many terms", use S∞ − Sₙ = u₁rⁿ ÷ (1 − r), solve with logs or a GDC table, and give the least whole number, checking both sides of it.
In context, draw the situation before writing the series, and decide which part is not in the geometric pattern (the first drop, a first payment).

10Try it

Marks in brackets. Q1 to Q4 are Paper 1 style (no calculator). Q5 and Q6 are Paper 2 style.

Q1. Find the sum to infinity of 54 + 18 + 6 + …. 3 marks

Q2. The sum to infinity of a geometric series is three times its first term. Find the common ratio. 3 marks

Q3. Use a geometric series to write 0.121212… as a fraction in its lowest terms. 3 marks

Q4. Consider the series 2 + 2(3x − 1) + 2(3x − 1)² + ….

(a) Find the values of x for which the sum to infinity exists. 3 marks

(b) Given that the sum to infinity is 5, find x. 3 marks

Q5. A geometric sequence has first term 20 and common ratio 0.9.

(a) Find the sum to infinity. 2 marks

(b) Find the least value of n for which S∞ − Sₙ < 1. 4 marks

Q6. A ball is dropped from a height of 5 m. Each time it hits the ground it rebounds to 80% of the height from which it fell. Find the total vertical distance it travels before coming to rest. 4 marks

11In one breath

An infinite geometric series has a sum only when its terms shrink fast enough, which means |r| < 1, that is −1 < r < 1 with both ends excluded: then rⁿ tends to 0, the partial sums close in on a limit, and that limit is S∞ = u₁ ÷ (1 − r), from the booklet. If |r| ≥ 1 there is no sum to infinity, whatever the formula says. Always find r first and write |r| < 1; bracket a negative r so that 1 − (−0.5) becomes 1.5. The formula runs backwards to find r or u₁, sometimes through a quadratic whose roots must each be tested. When r contains x, |r| < 1 opens into a double inequality that gives an interval for x, and any x found later must lie inside it. The gap after n terms is u₁rⁿ ÷ (1 − r), solved with logs for "how many terms". Recurring decimals and bouncing balls are geometric series in context; in the bouncing ball, the first drop happens once and every later height twice.

Answers

Q1. r = 18 ÷ 54 = 1/3, and |1/3| < 1. S∞ = 54 ÷ (1 − 1/3) = 54 ÷ (2/3) = 81. A1 for r = 1/3, M1 for substituting into u₁ ÷ (1 − r), A1 for 81.

Q2. u₁ ÷ (1 − r) = 3u₁. Divide by u₁ (which is not zero): 1 ÷ (1 − r) = 3, so 1 − r = 1/3 and r = 2/3. This satisfies |r| < 1. M1 for setting S∞ = 3u₁ with the formula, M1 for cancelling u₁ and rearranging, A1 for 2/3.

Q3. 0.121212… = 0.12 + 0.0012 + 0.000012 + …, with u₁ = 0.12 and r = 0.01. S∞ = 0.12 ÷ 0.99 = 12/99 = 4/33. M1 for writing the decimal as a geometric series with u₁ and r identified, A1 for 12/99, A1 for 4/33 in lowest terms.

Q4. (a) r = 3x − 1, so |3x − 1| < 1, so −1 < 3x − 1 < 1, so 0 < 3x < 2, so 0 < x < 2/3. (b) 2 ÷ (1 − (3x − 1)) = 5, so 2 ÷ (2 − 3x) = 5, so 2 − 3x = 2/5, so 3x = 8/5 and x = 8/15. Since 0 < 8/15 < 2/3, this is valid. (a) M1 for r = 3x − 1 and |r| < 1, M1 for the double inequality, A1 for 0 < x < 2/3 with strict signs. (b) M1 for S∞ in terms of x set equal to 5, A1 for x = 8/15, R1 for checking it lies in the interval from (a).

Q5. (a) S∞ = 20 ÷ (1 − 0.9) = 200. (b) S∞ − Sₙ = 20 × 0.9ⁿ ÷ 0.1 = 200 × 0.9ⁿ. So 200 × 0.9ⁿ < 1, 0.9ⁿ < 0.005, n > ln 0.005 ÷ ln 0.9 = 50.3, so n = 51. Check: n = 50 gives 1.03, n = 51 gives 0.928. (a) M1 A1. (b) M1 for an expression for S∞ − Sₙ, M1 for the inequality solved with logs or a GDC table, A1 for 50.3 or the table values either side, A1 for 51. An answer of 50 loses the final A1.

Q6. Rebound heights: 4, 3.2, 2.56, …, with u₁ = 4 and r = 0.8. Total = 5 + 2 × 4 ÷ (1 − 0.8) = 5 + 2 × 20 = 45 m. M1 for identifying the rebound series with u₁ = 4 and r = 0.8, M1 for S∞ = 20, M1 for 5 + 2 × S∞ (first drop once, rebounds twice), A1 for 45 m. 2 × 25 = 50 m (doubling the first drop) scores M1 M1 M0 A0.

Educerie · written from the published IB Diploma Programme Mathematics: analysis and approaches guide, first assessment 2021, section 1.8 Sum of infinite geometric sequences. Original text, examples and questions. Diagrams drawn by Educerie. Last reviewed 25 September 2026.

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