IB Mathematics: Analysis and Approaches SL/HL — Topic 1 Number and algebra
Infinite Geometric Series
Understand when an infinite geometric series converges, calculate its sum to infinity and use the result in algebraic and modelling problems.
Learning goal
Decide whether an infinite geometric series converges and calculate or use its sum to infinity.
Syllabus link
AA SL/HL: SL 1.8. Sum of infinite convergent geometric sequences; use of \,\(|r|<1\) and modulus notation.
Big idea
Infinitely many non-zero terms can have a finite total when their magnitudes shrink quickly enough.
Key relationship
For \,\(|r|<1\), \(S_\infty=\dfrac{u_1}{1-r}\).
Continuation from SL 1.3. In SL 1.3, the finite geometric sum is \(S_n=\dfrac{u_1(1-r^n)}{1-r}\). SL 1.8 asks what happens to that partial sum as the number of terms becomes arbitrarily large.
From finite partial sums to an infinite series
Consider the geometric series
\[1+\frac12+\frac14+\frac18+\cdots.\]Its first partial sums are
\[S_1=1,\quad S_2=1.5,\quad S_3=1.75,\quad S_4=1.875,\quad S_5=1.9375,\ldots\]Each additional term moves the partial sum closer to 2. No finite partial sum is exactly 2, but the partial sums can be made as close to 2 as we wish.
What “sum to infinity” means
Writing \(S_\infty=2\) does not mean that a finite partial sum eventually reaches 2, and there is no “last term”. It means the sequence of partial sums \(S_1,S_2,S_3,\ldots\) converges to 2.
When does an infinite geometric series converge?
A geometric term has the form \(u_n=u_1r^{n-1}\). For a non-zero geometric series to settle to a finite total, the terms being added must approach zero. This happens exactly when the magnitude of the common ratio is less than 1.
Worked example 1: decide whether each series converges
- \(8+4+2+1+\cdots\): \(r=\tfrac12\), so it converges.
- \(15-12+9.6-7.68+\cdots\): \(r=-0.8\), so it converges.
- \(3+3.6+4.32+\cdots\): \(r=1.2\), so it diverges.
- \(5-5+5-5+\cdots\): \(r=-1\), so the partial sums alternate between 5 and 0 and do not converge.
The modulus is essential
The condition is \(|r|<1\), not merely \(r<1\). For example, \(r=-2\) satisfies \(r<1\), but \(|r|=2>1\), so the term magnitudes grow.
The sum-to-infinity formula
Once convergence has been checked, the sum to infinity is
Worked example 2: positive common ratio
Find the sum to infinity of \(12+6+3+1.5+\cdots\).
Here \(u_1=12\) and \(r=\tfrac12\). Since \(|r|<1\),
\[S_\infty=\frac{12}{1-\frac12}=\boxed{24}.\]Why the formula works
From SL 1.3,
\[S_n=\frac{u_1(1-r^n)}{1-r}.\]If \(|r|<1\), then \(r^n\to0\) as \(n\to\infty\). Therefore
\[S_n\to\frac{u_1(1-0)}{1-r}=\frac{u_1}{1-r}.\]Converging geometric series and \(S_\infty\)
This tutorial reinforces when an infinite geometric series converges, why \(|r|<1\) matters and how to calculate the sum to infinity.
Visualising how partial sums approach the limit
| Ratio | Behaviour of terms | Behaviour of partial sums |
|---|---|---|
\(0| Same sign; magnitudes decrease to 0 | Approach the limit from one side | |
\(-1| Signs alternate; magnitudes decrease to 0 | Oscillate above and below the limit, with smaller jumps | |
| \(|r|=1\) | Terms do not shrink to 0 | Do not settle to one finite limit |
| \(|r|>1\) | Magnitudes grow | Do not settle to one finite limit |
Negative ratios: convergence with alternating signs
A negative common ratio does not prevent convergence. What matters is its magnitude.
Worked example 3: alternating convergent series
Find the sum to infinity of \(20-10+5-2.5+\cdots\).
Here \(u_1=20\) and \(r=-\tfrac12\). Since \(|r|=\tfrac12<1\),
\[S_\infty=\frac{20}{1-(-\frac12)}=\frac{40}{3}\approx\boxed{13.3}.\]The partial sums \(20,10,15,12.5,13.75,\ldots\) alternate around the limiting value.
Finding unknowns from \(S_\infty\)
The sum-to-infinity formula is an equation. If two of \(S_\infty\), \(u_1\) and \(r\) are known, the third can often be found directly.
Worked example 4: find the common ratio
An infinite geometric series has first term 12 and sum to infinity 30.
\[30=\frac{12}{1-r}\Rightarrow30(1-r)=12\Rightarrow1-r=0.4\Rightarrow\boxed{r=0.6}.\]The check \(|0.6|<1\) confirms that the infinite sum is valid.
Worked example 5: the sum is a multiple of the first term
If \(S_\infty=5u_1\), then for \(u_1\ne0\),
\[5u_1=\frac{u_1}{1-r}\Rightarrow5=\frac1{1-r}\Rightarrow\boxed{r=\frac45}.\]Parameters and the convergence condition
IB questions often hide the ratio inside an expression containing a parameter. First identify \(r\), then impose \(|r|<1\).
Worked example 6: find all parameter values for convergence
If \(r=2k-3\), then
\[|2k-3|<1\Rightarrow-1<2k-3<1\Rightarrow2<2k<4\Rightarrow\boxed{1Worked example 7: parameter and sum to infinity
Consider \(6+6(2a-1)+6(2a-1)^2+\cdots\). The ratio is \(r=2a-1\).
\[|2a-1|<1\Rightarrow\boxed{0Solve an inequality, not an equation
The condition \(|r|<1\) usually produces an interval of parameter values rather than isolated solutions.
Extension: how close is a finite partial sum to \(S_\infty\)?
Extension beyond the core SL 1.8 statement
The syllabus requires infinite convergent geometric series, \(|r|<1\) and the sum to infinity. The remainder is included because it follows naturally and appears in useful approximation-style questions; treat it as a consequence, not another formula to memorise.
The remainder after the first \(n\) terms is
\[R_n=S_\infty-S_n.\]Using the finite- and infinite-sum formulae,
\[R_n=\frac{u_1r^n}{1-r}=\frac{u_{n+1}}{1-r}.\]Worked example 8: how many terms are enough?
For \(100+60+36+\cdots\), find the least \(n\) for which \(S_n\) is within 0.1 of \(S_\infty\).
Here \(u_1=100\), \(r=0.6\), so
\[R_n=\frac{100(0.6)^n}{1-0.6}=250(0.6)^n.\]Require \(250(0.6)^n<0.1\). A GDC or logarithms gives \(n>15.3\ldots\), so the least integer is \(\boxed{16}\).
Application: an infinite bouncing-ball distance
If a ball is dropped from height \(H\) and each rebound reaches a fixed fraction \(r\) of the previous maximum height, with \(0 A ball is dropped from 3 m and reaches 70% of its previous maximum height after each bounce. The first rebound is \(h_1=3(0.70)=2.1\) m. The rebound-height sum is Therefore the total vertical distance is
How to count the distance
Worked example 9: total distance over all time
Recurring decimals as infinite geometric series
Some recurring decimals can be split into geometric blocks and summed exactly.
Worked example 10: convert a recurring decimal to a fraction
\[0.272727\ldots=0.27+0.0027+0.000027+\cdots.\]This has \(u_1=\tfrac{27}{100}\) and \(r=\tfrac1{100}\), so
\[S_\infty=\frac{27/100}{1-1/100}=\frac{27}{99}=\boxed{\frac3{11}}.\]Conceptual point
An infinite process does not automatically produce an infinite answer. What matters is how quickly the terms shrink.
Practice and answers
For every infinite-sum calculation, explicitly check the convergence condition first.
- State whether each ratio gives a convergent infinite geometric series: \(0.7,-0.7,1,-1,1.03,-0.92\).
- Find the sum to infinity of \(18+9+4.5+\cdots\).
- Find the sum to infinity of \(24-8+\frac83-\frac89+\cdots\).
- The first term is 15 and \(r=-0.4\). Find \(S_\infty\).
- A convergent geometric series has \(u_1=14\) and \(S_\infty=35\). Find \(r\).
- The sum to infinity is four times the first term. Find \(r\).
- Find all real \(k\) for which \(r=3k-2\) gives convergence.
- For \(\sum_{j=0}^{\infty}8(0.25)^j\), write the first three terms and find the sum to infinity.
- Find \(\sum_{k=2}^{\infty}5(0.4)^k\).
- For \(10+4+1.6+\cdots\), find \(S_5\), \(S_\infty\), and the exact value of \(S_\infty-S_5\).
- Extension: for \(50+30+18+\cdots\), find the least \(n\) such that \(S_n\) is within 0.05 of \(S_\infty\).
- A ball is dropped from 4.5 m and rebounds to 60% of its previous maximum height. Find the total vertical distance in the infinite model.
- Express \(0.454545\ldots\) as an exact fraction using an infinite geometric series.
- Explain why “\(r=-1.4<1\), so the series converges” is incorrect.
- A convergent series has first term \(a>0\) and \(S_\infty=3a\). Find \(r\) and \(u_2\) in terms of \(a\).
Answers 1–5
- Converges: \(0.7,-0.7,-0.92\). Diverges: \(1,-1,1.03\).
- \(u_1=18,r=\tfrac12\Rightarrow\boxed{36}\).
- \(r=-\tfrac13\Rightarrow S_\infty=24/(1+1/3)=\boxed{18}\).
- \(15/(1+0.4)=75/7\approx\boxed{10.7}\).
- \(35=14/(1-r)\Rightarrow\boxed{r=0.6}\).
Answers 6–10
- \(4u_1=u_1/(1-r)\Rightarrow\boxed{r=3/4}\).
- \(|3k-2|<1\Rightarrow\boxed{1/3
- First three terms: \(8,2,0.5\); \(S_\infty=8/(1-0.25)=\boxed{32/3}\).
- First displayed term is \(5(0.4)^2=0.8\), so the sum is \(0.8/0.6=\boxed{4/3}\).
- \(S_5=16.496\), \(S_\infty=50/3\), and \(S_\infty-S_5=\boxed{64/375}\).
Answers 11–15
- \(R_n=125(0.6)^n<0.05\Rightarrow\boxed{n=16}\).
- First rebound is 2.7 m; rebound sum is 6.75 m; total \(=4.5+2(6.75)=\boxed{18.0\text{ m}}\).
- \(0.45+0.0045+\cdots\Rightarrow\boxed{5/11}\).
- The condition is \(|r|<1\); here \(|-1.4|=1.4>1\), so it diverges.
- \(3a=a/(1-r)\Rightarrow r=2/3\), hence \(\boxed{u_2=2a/3}\).
Formula summary
| Convergence | \(|r|<1\) |
|---|---|
| Sum to infinity | \(S_\infty=\dfrac{u_1}{1-r}\), valid only for \(|r|<1\) |
| Finite partial sum | \(S_n=\dfrac{u_1(1-r^n)}{1-r}\) |
| Remainder (extension) | \(R_n=S_\infty-S_n=\dfrac{u_1r^n}{1-r}=\dfrac{u_{n+1}}{1-r}\) |
| Parameter test | Identify \(r\), then solve \(|r|<1\) |
| Bouncing-ball model | Initial drop once; each rebound height twice |
Checklist
- Explain \(S_\infty\) as a limit of partial sums.
- State and use \(|r|<1\).
- Recognise that negative ratios can converge.
- Derive and use the sum-to-infinity formula.
- Find \(r\), \(u_1\) or a parameter from \(S_\infty\).
- Read infinite sums in sigma notation.
- Compare \(S_n\) and \(S_\infty\).
- Use a remainder as an approximation error (extension).
- Model total bouncing-ball distance.
- Convert recurring decimals to fractions.
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