The big idea: A geometric series goes on forever. But sometimes, all those infinite terms add up to a finite number.__LINEBREAK__The answer depends on one thing: is each term getting smaller and smaller toward zero?
The rule
| Common ratio r | What happens to the terms? | Does S∞ exist? |
|---|---|---|
| r = 2 (e.g. 2, 4, 8, 16, ...) | Terms grow bigger and bigger | ❌ No |
| r = 1 (e.g. 5, 5, 5, 5, ...) | Terms stay the same forever | ❌ No |
| r = −1 (e.g. 3, −3, 3, −3, ...) | Terms flip sign, never shrink | ❌ No |
| r = 0.5 (e.g. 8, 4, 2, 1, ...) | Terms shrink toward zero | ✅ Yes |
| r = −0.5 (e.g. 8, −4, 2, −1, ...) | Terms shrink in size (ignore sign) | ✅ Yes |
The S∞ condition: S∞ exists only when |r| < 1__LINEBREAK___|r| means the size of r, ignoring any negative sign. So r = −0.7 → |r| = 0.7 → |r| < 1 ✓__LINEBREAK__If |r| ≥ 1 → write "S∞ does not exist". Never calculate it. You will lose marks if you do.
Practice: does S∞ exist?
Series: 3 + 6 + 12 + 24 + ...
r = 6 ÷ 3 = 2 → |r| = 2 ≥ 1__LINEBREAK___S∞ does not exist. The terms keep growing.
Series: 10 + 5 + 2.5 + 1.25 + ...
r = 5 ÷ 10 = 0.5 → |r| = 0.5 < 1 ✓__LINEBREAK___S∞ exists.
Series: 6 − 4 + 8/3 − 16/9 + ...
r = −4 ÷ 6 = −2/3 → |r| = 2/3 < 1 ✓__LINEBREAK___S∞ exists. Negative r is fine as long as |r| < 1.
Series: 5 + 5 + 5 + 5 + ...
r = 5 ÷ 5 = 1 → |r| = 1 → NOT less than 1__LINEBREAK___S∞ does not exist. The terms never shrink.
The formula: Once you confirm |r| < 1, use:__LINEBREAK___S∞ = u₁ ÷ (1 − r)__LINEBREAK__You always need two things: the first term u₁, and the common ratio r.
How to remember this (not on the formula sheet): S∞ is not given in the IB formula booklet — only Sₙ is.__LINEBREAK__But you can build it from the formula you do have on your sheet:__LINEBREAK___Sₙ = u₁(1 − rⁿ) ÷ (1 − r)__LINEBREAK__When |r| < 1, rⁿ shrinks to zero as n → ∞. So just replace rⁿ with 0:__LINEBREAK___S∞ = u₁(1 − 0) ÷ (1 − r) = u₁ ÷ (1 − r)__LINEBREAK__You are not memorising a new formula — you are crossing out rⁿ from the one already on your sheet.
Type 1 — Find S∞ from a sequence
The sequence 8, 4, 2, 1, ... goes on forever. Does S∞ exist? If so, find it.
Step by step
- Find r: r = 4 ÷ 8 = 0.5
- Check: |0.5| < 1 ✓ — S∞ exists
- Write the formula: S∞ = u₁ ÷ (1 − r)
- Substitute: S∞ = 8 ÷ (1 − 0.5) = 8 ÷ 0.5
- S∞ = 16
Final answer
S∞ = 16
Type 2 — Find u₁ given S∞ and r
A geometric series has r = 0.4 and S∞ = 25. Find the first term u₁.
Step by step
- Check: |0.4| < 1 ✓ (we're told S∞ = 25, so it already exists)
- Write the formula: S∞ = u₁ ÷ (1 − r)
- Substitute: 25 = u₁ ÷ (1 − 0.4)
- Simplify the bracket: 25 = u₁ ÷ 0.6
- Multiply both sides by 0.6: u₁ = 25 × 0.6
- u₁ = 15
Final answer
u₁ = 15
Type 3 — Find r given S∞ and u₁
A geometric series has u₁ = 12 and S∞ = 20. Find r.
Step by step
- Write the formula: S∞ = u₁ ÷ (1 − r)
- Substitute: 20 = 12 ÷ (1 − r)
- Rearrange: 1 − r = 12 ÷ 20 = 0.6
- r = 1 − 0.6
- r = 0.4
Final answer
r = 0.4
Rearranging the formula: The formula S∞ = u₁ ÷ (1 − r) has three variables.__LINEBREAK__If you know any two, you can always find the third:__LINEBREAK__- Find S∞: substitute u₁ and r directly - Find u₁: multiply both sides by (1 − r) - Find r: substitute S∞ and u₁, then solve 1 − r = u₁ ÷ S∞