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Learn7357Paper 1 · PureSequences and seriesSum to infinity
Sequences and series · Subtopic 4 of 5

Sum to Infinity

How adding infinitely many numbers can give a finite answer, the condition ∣r∣<1|r|<1 that makes it happen, and how to use S∞=a1−rS_\infty=\dfrac{a}{1-r} for recurring decimals, bouncing balls and exam problems.

Cambridge 9709 Paper 1Edexcel YMA01 WMA12Edexcel 9MA0 Paper 1Edexcel 9MA0 Paper 2AQA 7357 Paper 1AQA 7357 Paper 2≈ 35 minRevision 6 · updated 2026-10-27
Section 1 of 14

Learning objectives

  • Explain why a geometric series converges when ∣r∣<1|r|<1 and diverges otherwise.
  • Use S∞=a1−rS_\infty=\dfrac{a}{1-r}.
  • Write a recurring decimal as a fraction using a sum to infinity.
  • Solve problems combining S∞S_\infty with other information about a GP.
Pause and think

Walk halfway to a wall, then half the remaining distance, then half again, forever. You walk 12+14+18+…\tfrac12+\tfrac14+\tfrac18+\dots of the distance. Do you ever travel further than the wall?

Section 2 of 14

Core idea

Start hereZeno's impossible walk

To cross a room you must first go half way, then half the remaining distance, then half again... infinitely many steps. The ancient philosopher Zeno argued you could never arrive. But 12+14+18+⋯\frac12+\frac14+\frac18+\cdots adds up to exactly 11: infinitely many pieces, finite total. When does an infinite sum settle down, and to what?

The big idea

If −1<r<1-1<r<1, then rn→0r^n\to0 and the GP's sum approaches S∞=a1−rS_\infty=\dfrac a{1-r}. If ∣r∣≥1|r|\ge1, the sum does not settle: the series diverges.

1. Seeing it

Halving each time fills exactly one whole
1/21/41/8
A square of area 11: take half, then half of what is left, and so on. The pieces 12+14+18+…\tfrac12+\tfrac14+\tfrac18+\dots fill the square but never spill over, so their total is exactly 11.
Partial sums of 1 + ½ + ¼ + … approach 2
24680.511.520number of terms nSₙS∞ = 2
S∞=a1−r=11−12=2S_\infty=\dfrac{a}{1-r}=\dfrac{1}{1-\tfrac12}=2.

2. The formula

From SnS_n to S∞S_\infty
  1. Sn=a(1−rn)1−rS_n=\frac{a(1-r^n)}{1-r}
    Finite sum.
  2. ∣r∣<1⇒rn→0 as n→∞|r|<1\Rightarrow r^n\to0\text{ as }n\to\infty
    E.g. 0.520≈10−60.5^{20}\approx10^{-6}: repeated multiplication by a fraction shrinks.
  3. S∞=a(1−0)1−r=a1−rS_\infty=\frac{a(1-0)}{1-r}=\frac a{1-r}
    Replace rnr^n by its limit.
Valid only for ∣r∣<1|r|<1.
How r affects convergence
24681012246810120number of terms nSₙ
With a=1a=1: the partial sums (solid) rise towards the limit 11−r\dfrac{1}{1-r} (dashed). The closer rr is to 11, the bigger the limit and the slower the approach.
Common misconceptionUsing S∞S_\infty when ∣r∣≥1|r|\ge1 1+2+4+⋯1+2+4+\cdots has no sum to infinity. The formula would give 11−2=−1\frac1{1-2}=-1: nonsense. Always check ∣r∣<1|r|<1 first.
Worked exampleRecurring decimal as a fraction

Write 0.27‾=0.272727…0.\overline{27}=0.272727\ldots as a fraction.

Goal
Simplest fraction.
Strategy
See it as a GP.
  1. 0.27+0.0027+0.000027+⋯0.27+0.0027+0.000027+\cdots
    Each term is 1100\frac1{100} of the last.
  2. a=0.27, r=0.01a=0.27,\ r=0.01
    ∣r∣<1|r|<1 ✓.
  3. S∞=0.270.99=2799=311S_\infty=\frac{0.27}{0.99}=\frac{27}{99}=\frac3{11}
    Formula; simplify.
Answer311\frac3{11}.
✓ Check3÷11=0.2727…3\div11=0.2727\ldots ✓.
Worked exampleCondition for convergence

For which xx does 1+2x+4x2+⋯1+2x+4x^2+\cdots converge? Find its sum.

Goal
Interval of xx; S∞S_\infty.
Strategy
r=2xr=2x; require ∣r∣<1|r|<1.
  1. ∣2x∣<1⇒−12<x<12|2x|<1\Rightarrow-\tfrac12<x<\tfrac12
    Convergence condition.
  2. S∞=11−2xS_\infty=\frac1{1-2x}
    a=1a=1.
Answer−12<x<12-\frac12<x<\frac12; sum 11−2x\frac1{1-2x}.
✓ Checkx=14x=\frac14: 1+12+14+⋯=2=11−121+\frac12+\frac14+\cdots=2=\frac1{1-\frac12} ✓.
Worked exampleBouncing ball

A ball dropped from 1010 m rebounds to 60%60\% of its height each time. Total distance travelled?

Goal
Infinite total.
Strategy
First drop, then pairs of up-and-down forming a GP.
  1. 10+2(6+3.6+2.16+⋯ )10+2(6+3.6+2.16+\cdots)
    Each rebound is travelled up and down.
  2. 6+3.6+⋯=61−0.6=156+3.6+\cdots=\frac6{1-0.6}=15
    a=6a=6, r=0.6r=0.6.
  3. 10+30=40 m10+30=40\text{ m}
    Total.
Answer4040 m.
✓ CheckFinite, though the bounces never stop in theory ✓.
Pause and think

S∞=3×S_\infty=3\times the first term. Find rr.

Summary

  • S∞=a1−rS_\infty=\frac a{1-r}, only for ∣r∣<1|r|<1.
  • Convergence conditions give inequalities.
Key words
Converge
Approach a finite limit.
Diverge
Not settle to a finite value.
HistoryZeno's paradox. Around 450 BC, Zeno of Elea argued that Achilles could never catch a tortoise: by the time he reaches where it was, it has moved on a little, and so on forever. For two thousand years this puzzled thinkers. The sum to infinity resolves it: infinitely many shorter and shorter time intervals add up to a finite time.
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