A geometric series with |r| < 1 has a fixed total as the number of terms grows without limit. That total is S∞ = a ÷ (1 − r), and it exists only when −1 < r < 1.
This lesson is part of progressions. It builds on terms and sums of geometric progressions.
Why does the condition matter?
When |r| < 1, each term is smaller than the one before, so the added amounts shrink and the total approaches a limit. When |r| ≥ 1, the terms do not shrink, so the total grows without settling.
The formula S∞ = a ÷ (1 − r) still gives a number if r = 2, but that number means nothing. The condition must be checked first.
Worked example: one series that settles, one that does not
Here is an original contrast. Series A is 12 + 4 + 4/3 + … and series B is 12 + 24 + 48 + …
Series A. r = 4 ÷ 12 = 1/3. Since |1/3| < 1, the sum to infinity exists.
S∞ = 12 ÷ (1 − 1/3) = 12 ÷ (2/3) = 12 × 3/2 = 18.
Series B. r = 24 ÷ 12 = 2. Since |2| ≥ 1, the terms grow and no sum to infinity exists. Writing 12 ÷ (1 − 2) = −12 would be wrong, because the series only gets larger.
The running totals of series A are 12, 16, 17.33 and 17.78, moving towards 18.
Finding r from the sum
Sometimes the sum is given. A geometric series has a = 10 and S∞ = 40. Find r.
- Write 10 ÷ (1 − r) = 40.
- Rearrange: 1 − r = 10 ÷ 40 = 1/4.
- Solve: r = 3/4.
- Check: |3/4| < 1, so the sum exists, and 10 ÷ (1/4) = 40.
The check in step 4 confirms that the answer obeys the condition. If you had found r = 2, you would know something had gone wrong.
A range-of-values question
A geometric series has first term 9 and second term 3x. Find the range of x for which the sum to infinity exists.
The ratio is r = 3x ÷ 9 = x/3.
The condition is −1 < x/3 < 1, so −3 < x < 3.
Test a value inside the range. For x = 1.5, r = 0.5 and S∞ = 9 ÷ 0.5 = 18. The sum exists, as expected.
The mistake that costs marks
The common slip is to apply a ÷ (1 − r) without checking r. For Series B this gives a negative number for a series of growing positive terms, which should alert you.
| Step | Wrong | Right |
|---|---|---|
| Find r | (skipped) | r = 2 |
| Check that r is between −1 and 1 | (skipped) | Fails |
| Conclusion | S∞ = −12 | No sum to infinity |
Write the condition as a line in your working. Even when it passes, the line shows that you checked.
A recurring decimal
Write 0.444… as 0.4 + 0.04 + 0.004 + … The first term is 0.4 and r = 0.1, so S∞ = 0.4 ÷ (1 − 0.1) = 0.4 ÷ 0.9 = 4/9.
Check yourself
A geometric series has a = 20 and S∞ = 25. Find r, and state the second term.
Answer
20 ÷ (1 − r) = 25, so 1 − r = 20 ÷ 25 = 4/5, and r = 1/5.
|1/5| < 1, so the sum exists.
The second term is ar = 20 × 1/5 = 4.
Check: 20 ÷ (4/5) = 25.
What to study next
The sum to infinity appears in some word problems, such as a bouncing ball or a repeated percentage decrease. Continue with translating payment or growth patterns into progressions, and then use the practice set.
If you want a teacher to give you series that tempt you to skip the check, see online one-to-one Additional Mathematics tuition.