Series, Sigma Notation & Convergence — Grade 12 Maths (CAPS)

Sum formulae for arithmetic and geometric series, reading and writing sigma notation, and the sum to infinity — including the values of the ratio for which a series converges.

A sequence is a list of terms; a series is what you get when you add them. Sₙ means the sum of the first n terms.

1The formulae

Arithmetic
Tₙ = a + (n − 1)d · Sₙ = n/2[2a + (n − 1)d]

Geometric
Tₙ = ar^{n − 1} · Sₙ = (a(rⁿ − 1))/(r − 1), r ≠ 1

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If you already know the last term ℓ, the arithmetic sum is quicker as Sₙ = n/2(a + ℓ).

Worked ExampleTₙ = 2n + 4. Determine S₁₉

1
a = T₁ = 2(1) + 4 = 6, and d = 2.
The coefficient of n is the common difference.
2
S₁₉ = 19/2[2(6) + 18(2)].
Substitute into the arithmetic sum formula.
3
= 19/2[12 + 36] = 19/2(48).
4
= 456.

2Sigma notation

∑_n=1^k Tₙ = T₁ + T₂ + … + Tₖ

The letter below the ∑ is the counter and its starting value; the number on top is where it stops. Number of terms = top − bottom + 1.

Worked ExampleEvaluate ∑_n=1²⁰ (3n − 1)

1
This is arithmetic: a = 3(1) − 1 = 2, d = 3, n = 20.
20 − 1 + 1 = 20 terms.
2
S₂₀ = 20/2[2(2) + 19(3)] = 10[4 + 57].
3
= 10(61) = 610.

⚠️

∑_n=3¹⁰ has 8 terms, not 7 — it is 10 − 3 + 1. Getting the number of terms wrong loses every mark that follows.

3Writing a series in sigma notation

Worked ExampleWrite 2 + 5 + 8 + … + 59 in sigma notation

1
Arithmetic with a = 2, d = 3, so Tₙ = 3n − 1.
a + (n − 1)d = 2 + 3n − 3.
2
Find where it stops: 3n − 1 = 59 → n = 20.
3
∑_n=1²⁰ (3n − 1).

4Sum to infinity and convergence

S_∞ = a/(1 − r) only when −1 < r < 1

If |r| < 1 the terms shrink towards zero, so the total settles on a fixed value — the series converges. If |r| ≥ 1 the terms do NOT approach zero, so the sum never settles and the series diverges. (They need not grow: at r = 1 every term is the same, and at r = −1 they just flip between +a and −a.)

Worked ExampleDetermine S_∞ for 8 + 4 + 2 + 1 + …

1
r = 4/8 = 1/2, and |r| < 1, so it converges.
Always state this check.
2
S_∞ = 8/(1 − 1/2) = 8/(1/2).
3
= 16.

Worked ExampleFor which values of x does ∑ 3(2x)^{n − 1} converge?

1
Here r = 2x.
The base of the power is the ratio.
2
Need −1 < 2x < 1.
The convergence condition.
3
−1/2 < x < 1/2.
Divide throughout by 2.

5Where the sum formulae come from

The ATP asks for the derivation, not just the use, and it is examinable. Both derivations are short, and both turn on one trick.

Worked ExampleDerive Sₙ = n/2 [2a + (n − 1)d] for an arithmetic series.

1
Write the sum forwards: Sₙ = a + (a + d) + (a + 2d) + … + (l − d) + l
l is the last term, a + (n − 1)d.
2
Write the same sum backwards: Sₙ = l + (l − d) + (l − 2d) + … + (a + d) + a
This is the trick — the whole derivation rests on it.
3
Add the two lines term by term: 2Sₙ = (a + l) + (a + l) + … + (a + l)
Every pair adds to the same total, because the d's cancel.
4
There are n such pairs, so 2Sₙ = n(a + l)
5
Sₙ = n/2 (a + l)
The 'last term' form — use it when you know the last term.
6
Substituting l = a + (n − 1)d gives Sₙ = n/2 [2a + (n − 1)d]
The standard form — use it when you know a, d and n.

Worked ExampleDerive Sₙ = (a(rⁿ − 1))/(r − 1) for a geometric series.

1
Sₙ = a + ar + ar² + … + arⁿ⁻¹
2
Multiply the whole line by r: rSₙ = ar + ar² + … + arⁿ⁻¹ + arⁿ
The trick here — almost every term now appears in both lines.
3
Subtract the first from the second: rSₙ − Sₙ = arⁿ − a
Everything in the middle cancels.
4
Sₙ(r − 1) = a(rⁿ − 1)
Factorise both sides.
5
Sₙ = (a(rⁿ − 1))/(r − 1), r ≠ 1
Division by r − 1 requires r ≠ 1 — state it.

The sum to infinity follows from the second result. When −1 < r < 1, rⁿ shrinks towards 0 as n grows, so the rⁿ term disappears:

S_∞ = a/(1 − r), for −1 < r < 1

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Remember the two tricks and you can rebuild either derivation: write it backwards and add for arithmetic, multiply by r and subtract for geometric.

Practice exercises

Work each one out, then click to reveal the answer.

📄 Worksheet ✅ Memo 📝 Exam paper 🗝️ Marking guide

1
The general term of an arithmetic sequence is Tₙ = 2n + 4. Determine T₁₀.
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2(10) + 4 = 24
2
Determine the sum of the first 19 terms of Tₙ = 2n + 4.
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a = 6, d = 2. S₁₉ = 19/2[12 + 36] = 456
3
Evaluate ∑_n=1²⁰ (3n − 1)
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a = 2, d = 3, n = 20. S = 10[4 + 57] = 610
4
How many terms are there in ∑_n=3¹⁰ Tₙ?
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10 − 3 + 1 = 8 terms
5
Write 2 + 5 + 8 + … + 59 in sigma notation.
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Tₙ = 3n − 1 and 3n − 1 = 59 gives n = 20, so ∑_n=1²⁰ (3n − 1)
6
Determine S_∞ for 8 + 4 + 2 + 1 + …
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r = 1/2, so S_∞ = 8/(1 − 1/2) = 16
7
Determine the values of x for which ∑ 3(2x)^{n − 1} converges.
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−1 < 2x < 1, so −1/2 < x < 1/2
8
Calculate S₁₀ for the geometric series 2 + 6 + 18 + …
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a = 2, r = 3. S₁₀ = (2(3¹⁰ − 1))/2 = 59 048
9
Derive the formula Sₙ = n/2 (a + l) for an arithmetic series.
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Write Sₙ forwards and backwards ✓; adding gives 2Sₙ = n(a + l) because each pair sums to a + l ✓✓; ∴ Sₙ = n/2 (a + l) ✓
10
Derive the formula for the sum of the first n terms of a geometric series.
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Sₙ = a + ar + … + arⁿ⁻¹; multiply by r ✓; subtract to get rSₙ − Sₙ = arⁿ − a ✓; factorise: Sₙ(r − 1) = a(rⁿ − 1) ✓; Sₙ = a(rⁿ − 1)/(r − 1), r ≠ 1 ✓
11
Explain why the sum to infinity only exists when −1 < r < 1.
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Only then does rⁿ approach 0 as n grows ✓, so the terms shrink and the total settles on a finite value ✓