2 + 6 + 18 + 54 + … Put a = 2 and r = 3 into a/(1 − r) and it returns −1. Every term of that series is positive, so −1 is not a sum of it. The condition |r| < 1 is not decoration.
The dots are the partial sums. The line is what the formula claims. They only belong together in the first view.
The sum of n terms, which you already have from 1.3, is
Sn = a(1 − rn) / (1 − r)
and everything turns on the rn. If |r| < 1 then rn shrinks to nothing as n grows: 0.610 is 0.006, and 0.650 is about eight trillionths. So the bracket closes on 1 and
S∞ = a / (1 − r)
That is the whole derivation, and it is worth being able to say, because it tells you exactly when the formula is allowed to exist: only when rn dies.
If |r| ≥ 1 the term rn does not die, so there is nothing to drop and no sum to find. The expression a/(1 − r) still evaluates, because arithmetic does not know what it is being asked. With a = 2 and r = 3 it returns −1, and the partial sums are 2, 8, 26, 80, and 59,048 by the tenth term. Nothing about −1 is a summary of those.
| n | r = 0.6: Sn | r = 3: Sn |
|---|---|---|
| 1 | 2.00 | 2 |
| 2 | 3.20 | 8 |
| 3 | 3.92 | 26 |
| 4 | 4.35 | 80 |
| 5 | 4.61 | 242 |
| 10 | 4.97 | 59 048 |
| ∞ | 5.00 | no sum exists |
Read the left column: the partial sums rise, slow down, and settle. They never reach 5, and 5 is the number they are heading for, which is the only honest meaning of an infinite sum. The right column does not settle on anything.
|r| < 1 means r lies strictly between −1 and 1. So:
The negative cases are the ones to be careful with. A negative r can still converge, and a student who tests r > 0 instead of |r| < 1 will throw away a perfectly good series.
Working backwards. A geometric series has first term 5 and sums to 20. Then
5/(1 − r) = 20, so 1 − r = 0.25, so r = 0.75
and 0.75 passes the test, so the answer stands. Always check the r you find against |r| < 1: a question can be built so that the algebra returns an r that the series cannot have.
Recurring decimals. 0.999… is the geometric series 0.9 + 0.09 + 0.009 + … with a = 0.9 and r = 0.1, so
S∞ = 0.9/0.9 = 1
exactly, not nearly. The two notations name the same number, and this is the shortest honest proof of it.
The formula is one keystroke, so the machine's real use is tabulating the partial sums. Seeing them approach a value, or run away from one, is what turns the condition from a rule into something you can check.
When you may use it. Analysis Paper 1 is non-calculator, and these answers are usually exact anyway: 2/0.4 is 5 and 0.9/0.9 is 1. Paper 2 allows the calculator and is where a tabulation is worth the time.
The mark people lose. Using the formula without checking |r| < 1. It is one inequality and it is often a mark in its own right on a "find the sum to infinity" question, because the examiner wants to see that you know when the formula applies. The other half of the same trap is testing r > 0: r = −0.6 converges perfectly well, to 1.25, and rejecting it loses the whole question.
1. A geometric series has a = 2 and r = 0.6. Find the sum to infinity.
2. For the same series, find the tenth partial sum, to 2 decimal places.
3. A series has first term 5 and sum to infinity 20. Find r.
4. Put a = 2 and r = 3 into a/(1 − r). What number comes out?
5. So what does that −1 mean?
1 markThe condition |r| < 1, checked or stated.
1 markThe formula, with a and r in the right places.
1 markThe answer, exact where it can be.
On a "for which values of r does this converge" question the inequality is the answer, and writing −1 < r < 1 earns it while r < 1 does not.
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