Key Idea: Add a geometric sequence forever and the total can settle to a finite number. It shows up on both papers, and the whole topic hinges on one convergence check.
♾️ The sum to infinity
- the first term
- the common ratio (next ÷ current)
S∞ exists only when |r| < 1, because the terms shrink toward 0. If |r| ≥ 1 the terms don't shrink, the total grows without limit, and there is no sum to infinity — give a finite sum instead. State |r| < 1 before you compute.
🔁 The exam variations
| Question type | What to do |
|---|---|
| Find S∞ from a series | Find r (next ÷ current), check |r| < 1, then S∞ = u₁/(1 − r). |
| Given S∞, find u₁ or r | Substitute into S∞ = u₁/(1 − r) and rearrange (denominator is 1 − r). |
| Least n within a tolerance | The gap is S∞ − Sₙ = u₁rⁿ/(1 − r); set it below the tolerance, round up. |
| |r| ≥ 1 — no S∞ | Use the finite sum Sₙ (e.g. n = 2m), and simplify with r²ᵐ = (r²)ᵐ. |
✏️ IB-style worked examples
IB-style question — find the sum to infinity
Find the sum to infinity of 18 + 6 + 2 + … .
Step by step:
Find r by dividing consecutive terms, and check it converges.
Substitute u₁ = 18 and r = ⅓ into the formula.
Finish.
S∞ = 27.
IB-style question — given S∞, find the first term
A geometric series has common ratio r = 0.4 and a sum to infinity of 45. Find the first term.
Step by step:
Write the formula and substitute what you know.
Work out the denominator (1 − r).
Multiply both sides by 0.6 to isolate u₁.
u₁ = 27.
IB-style question — total distance of a bouncing ball
A ball is dropped from 12 m and rebounds to ½ of its height each bounce, forever. Find the total distance it travels.
Step by step:
It drops 12 m once; then each rebound is travelled up and back down. The rebound heights 6, 3, 1.5, … are geometric (u₁ = 6, r = ½) — sum them to infinity.
Total = the drop, plus twice the rebound sum.
36 m. (Shortcut: total = x(1 + r)/(1 − r); with r = ⅔ that's the classic δ = 5x.)
Important: S∞ = u₁/(1 − r) only works when |r| < 1. If |r| ≥ 1 there is no sum to infinity — the question wants a finite sum Sₙ (often the first 2m terms). Always check |r| before reaching for S∞.
Tap each card to reveal the answer.
Sum to infinity of 8 + 4 + 2 + … r = ½, so S∞ = 8 / (1 − ½) = 16.
Does 5 + 10 + 20 + … have a sum to infinity? r = 2, and |r| ≥ 1 — no, the total grows without limit.
S∞ = 25 and u₁ = 10. Find r. 25 = 10/(1 − r) ⇒ 1 − r = 0.4 ⇒ r = 0.6.
What is the gap S∞ − Sₙ in terms of n? u₁rⁿ/(1 − r) — set it below the tolerance and round n up.
Simplify r²ᵐ when r = 3 3²ᵐ = (3²)ᵐ = 9ᵐ — the standard |r| ≥ 1 simplifying trick.
Exam tips
- S∞ exists ONLY when |r| < 1; then S∞ = u₁/(1 − r). State the check first.
- Find r as next ÷ current before doing anything else.
- Given S∞, rearrange for u₁ or r — the denominator is 1 − r, never r.
- Partial sums approach S∞: the gap is u₁rⁿ/(1 − r); set it below the tolerance and round up.
- If |r| ≥ 1 there is no S∞ — give a finite sum Sₙ, simplifying with r²ᵐ = (r²)ᵐ.