The big idea: Add a geometric sequence forever and the total settles to a finite number — but only when |r| < 1 (the terms shrink toward 0).
For example, 8 + 4 + 2 + 1 + … has r = ½, and the total is 16.
Converges only if |r| < 1: If |r| ≥ 1 the terms do not shrink, the total grows without limit, and there is no sum to infinity.
- the first term
- the common ratio, with |r| < 1
Plug into the formula: Once you know |r| < 1, just substitute into S∞ = u₁/(1 − r).
Find u₁ and r first if they are not given.
IB-style question — find S∞
Find the sum to infinity of 12 + 8 + 16/3 + … .
Step by step
- Find r — divide consecutive terms. Check |r| < 1.
- Substitute into the formula.
- Finish.
Final answer
S∞ = 36.
IB-style question — find r first
A geometric series has first term 20 and second term 5.
Find its sum to infinity.
Step by step
- Common ratio.
- Substitute.
- Finish.
Final answer
S∞ = 80/3 ≈ 26.7.
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Check convergence, then rearrange: Before using S∞, confirm |r| < 1.
Given the sum to infinity, you can rearrange S∞ = u₁/(1 − r) to find u₁ or r.
IB-style question — given S∞, find r
A geometric series has first term 10 and a sum to infinity of 25.
Find the common ratio.
Step by step
- Write the formula and substitute.
- Solve for (1 − r).
- Solve for r.
Final answer
r = 0.6.
Common mistakes
- Using S∞ when |r| ≥ 1 (there is no sum).
- Putting r in the denominator instead of (1 − r).
- Forgetting to check convergence first.
Do this instead
- Only use S∞ when |r| < 1.
- S∞ = u₁/(1 − r).
- State |r| < 1 before computing.
The gap to S∞ shrinks each term: A Paper 2 favourite: the partial sums creep toward S∞ — find the least n for which Sₙ is within a tiny tolerance of S∞ (i.e. the gap S∞ − Sₙ drops below a given amount).
IB-style question — least n within a tolerance (Paper 2)
A geometric series has first term 8 and common ratio 0.5.
Find the least value of n for which Sₙ is within 0.1 of the sum to infinity.
Step by step
- Find S∞ first.
- The gap left after n terms is the tail.
- Set the gap below the tolerance; read the GDC table (or logs) and round up.
Final answer
Least n = 8 (the gap is 0.125 at n = 7, and 0.0625 at n = 8).
GDC tip (Paper 2): Enter the gap 16(0.5)^X (or Sₙ itself) as Y₁ and read the table (2nd → GRAPH).
Round n up to the first whole number that is within the tolerance.
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No S∞ ⇒ give a finite sum: A harder twist: when |r| ≥ 1 the series has no sum to infinity, so a "sum" question must be finite — e.g. the first 2m terms, found by putting n = 2m into the Sₙ formula.
IB-style question — no S∞, sum of 2m terms
A geometric series has first term 2 and common ratio 3.
(a) Explain why it has no sum to infinity.
(b) Find the sum of the first 2m terms, in the form 9ᵐ − 1.
Step by step
- (a) The ratio is not less than 1 in size, so the terms grow and the total is unbounded.
- (b) Use the finite sum with n = 2m.
- Cancel the 2, then use 32m = (32)m = 9m.
Final answer
(a) |r| ≥ 1, so there is no sum to infinity. (b) S₂ₘ = 9ᵐ − 1.
Finite, not infinite: If |r| ≥ 1, reach for the Sₙ formula, never S∞.
The simplifying trick is almost always a power law: r^{2m} = (r²)ᵐ (here 3²ᵐ = 9ᵐ).
Let it bounce forever: If a ball bounces forever (|r| < 1), the total distance is the drop plus twice the sum to infinity of the rebounds: total = x + 2·S∞(rebounds).
IB-style question — total distance forever
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
- The rebounds (6, 3, 1.5, …) are geometric, u₁ = 6, r = ½. Sum to infinity:
- Total = the drop, plus twice the rebounds.
Final answer
36 m.
The neat shortcut: distance = x(1 + r)/(1 − r): For a ball dropped from x and rebounding to r forever:
total = x + 2·(rx/(1 − r)) = x(1 + r)/(1 − r).
For r = ½ that is 3x (here 3 × 12 = 36); for r = ⅔ it is 5x — the classic "show the distance is 5x" exam question.