Check geometric convergence and interpret a sum to infinity
Establish the ratio condition before using a limiting sum, handle alternating series and separate an idealised total from a finite-stage result.
Pure Mathematics 1 (Paper 1) · AS & A Level Mathematics 9709, 2026–2027. Use positive-integer binomial powers and positive-integer term indices. Written practice is self-checked.
Before you begin
- Use the finite GP sum and solve compound inequalities.
- Interpret a limiting value and include repeated journeys in a distance total.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which non-zero GP has a convergent sum to infinity?
What you will learn
- Use |r| < 1 to establish convergence of a non-zero geometric series.
- Find sums to infinity, remainders and parameter ranges, and apply them to a stated model.
The finite sums must approach one value
For a non-zero first term a, a GP has a convergent sum to infinity precisely when |r| < 1, meaning −1 < r < 1. Then rⁿ tends to zero and the finite formula Sₙ = a(1 − rⁿ)/(1 − r) approaches S∞ = a/(1 − r). A negative ratio can converge: its terms alternate with decreasing magnitude and its partial sums approach the same limit from alternate sides.
If r = 1, the non-zero terms never shrink; if r = −1, partial sums alternate rather than approaching one value; if |r| > 1, term magnitudes grow. A negative ratio is not automatically invalid, but a small positive first term does not guarantee convergence. The identically zero series with a = 0 is a trivial exception: every partial sum is zero for any ratio, so the stated non-zero-first-term condition matters.
A limit does not claim a finite process is already finished
For a = 8 and r = 1/2, S∞ = 16. After four terms the sum is 8 + 4 + 2 + 1 = 15 and the remaining limiting contribution is 1. In general, subtracting Sₙ from S∞ gives a rⁿ/(1 − r), whose sign can alternate for negative ratios. An infinite cutting or bouncing model is an idealisation; it does not promise a physical process can carry out infinitely many steps.
If the ratio contains a parameter, solve −1 < r < 1 before using the infinite-sum formula. For first term 9 and ratio 2k − 1, convergence requires 0 < k < 1, and S∞ = 9/[2(1 − k)] there. If a distance model contains a first drop followed by repeated upward and downward journeys, count the first drop once and each later bounce height twice; those heights alone are not the total distance.
Pause and explain
For first term 9 and ratio 2k − 1, which parameter range gives convergence?
Worked example
A GP has first term 9 and ratio 2k − 1. Find the convergence interval and k if its sum to infinity is 18.
Show the worked solution
- Convergence requires −1 < 2k − 1 < 1, which simplifies to 0 < k < 1.
- On that interval, S∞ = 9/[1 − (2k − 1)] = 9/[2(1 − k)].
- Set the expression equal to 18 to get 2(1 − k) = 1/2, hence k = 3/4. Its ratio is 1/2, satisfying the convergence condition.
Answer Converges for 0 < k < 1; S∞ = 18 when k = 3/4.
Try a topic test
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
A GP has a = 12 and r = 1/3. Find S∞ and the limiting contribution remaining after its first four terms.
Give me a hint
Check convergence, then subtract the finite sum or use the remainder formula.
Compare my reasoning
- Since |1/3| < 1, S∞ = 12/(1 − 1/3) = 18.
- The first-four sum is 18[1 − (1/3)⁴] = 160/9.
- The remaining limiting contribution is 18 − 160/9 = 2/9; it is not zero at a finite stage.
S∞ = 18; remainder after four terms 2/9.
Look for these in your work
- Establish convergence before applying the limiting formula.
- Distinguish the finite total from the remaining contribution.
For 10, −5, 5/2, …, find S∞, S₄ and S∞ − S₄ exactly.
Give me a hint
Keep the negative ratio inside its power and denominator brackets.
Compare my reasoning
- The ratio is −1/2, so the series converges and S∞ = 10/(3/2) = 20/3.
- S₄ = 10[1 − (−1/2)⁴]/(3/2) = 25/4.
- Subtract to get 20/3 − 25/4 = 5/12; the alternating remainder sign depends on which finite stage is chosen.
S∞ = 20/3; S₄ = 25/4; S∞ − S₄ = 5/12.
Look for these in your work
- Use the actual signed ratio in both formulae.
- Subtract the exact partial sum from the limit.
In an idealised vertical model, a ball first drops 12 m and then rebounds to 3/5 of its previous fall height after every impact. Find its total limiting distance, counting upward and downward travel.
Give me a hint
The rebound heights form a GP, but each is travelled twice after the initial drop.
Compare my reasoning
- The rebound heights start at 12(3/5) = 36/5 m, with ratio 3/5, which has magnitude below 1.
- Their infinite sum is (36/5)/(1 − 3/5) = 18 m.
- Total distance is the initial 12 m plus twice that height sum: 12 + 2(18) = 48 m. This is the limit of the supplied model, rather than a claim about infinitely many physical bounces.
48 m in the idealised limiting model.
Look for these in your work
- Identify the first rebound term separately from the first drop.
- Count each rebound's upward and downward legs and state the model limitation.
Common mistakes
- Check |r| < 1 before claiming a non-zero GP has a finite sum to infinity.
- A negative ratio can converge; a ratio of magnitude 1 does not.
- Count both upward and downward travel after a bounce, and distinguish an ideal limit from a finite stage.
What can you explain now?
Find the sum to infinity of 10, −5, 5/2, … and explain why it exists.
Compare with the explanation
20/3, because r = −1/2 has |r| < 1.
Use 10/[1 − (−1/2)] = 20/3; alternating signs do not prevent convergence when magnitudes shrink.
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Make it stick
Explain why a shrinking alternating GP can converge, derive its limiting sum from Sₙ and identify which finite stages or repeated journeys a model actually counts.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.