Sequences & Series | Ronny Best Mathematics

Foundation

What is a Sequence?

Foundation

A sequence is an ordered list of numbers, called terms, that follow a definite rule. The rule determines every term from its position in the list. Understanding sequences is the foundation of nearly all higher mathematics — from modelling population growth to describing oscillating electrical signals.

Types of Sequences

Every sequence has a general term formula, also written as the nth term, which gives any term directly from its position number n. The four most important types in the CBC curriculum are:

Type

General Term (aₙ)

Arithmetic

aₙ = a + (n − 1)d where d = common difference

Geometric

aₙ = arⁿ⁻¹ where r = common ratio

Quadratic (2nd difference)

aₙ = An² + Bn + C constant 2nd difference

Fibonacci

aₙ = aₙ₋₁ + aₙ₋₂ each term is sum of two before it

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CBC Key Concept: The KICD syllabus emphasises recognising a sequence's type before applying a formula. Always identify whether differences or ratios are constant before proceeding.

Worked Examples — Identifying and Finding Terms

Ex 1 Find the next two terms: 4, 7, 10, 13, ...

IdentifyDifferences: 7−4 = 3, 10−7 = 3, 13−10 = 3. Common difference d = 3 → Arithmetic sequence.

Next14th term: 13 + 3 = 16. 15th term: 16 + 3 = 19.

Next two terms: 16, 19

Ex 2 Find the next two terms: 2, 6, 18, 54, ...

IdentifyRatios: 6÷2 = 3, 18÷6 = 3, 54÷18 = 3. Common ratio r = 3 → Geometric sequence.

Next5th term: 54 × 3 = 162. 6th term: 162 × 3 = 486.

Next two terms: 162, 486

Ex 3 Find the nth term of: 1, 4, 9, 16, 25, ...

ObserveThese are perfect squares: 1² = 1, 2² = 4, 3² = 9, 4² = 16 ...

Verify1st differences: 3, 5, 7, 9 (odd numbers). 2nd differences: 2, 2, 2 → constant → quadratic.

Formulaaₙ = n². Check: a₅ = 5² = 25 ✔

aₙ = n²

Foundation

Notation & Terminology

Foundation

Before working with series, you must be fluent with the notation used throughout all levels of the CBC and KNEC examinations.

Standard Notation

General term:aₙ or Tₙ or uₙ — the nth term of the sequence

First term:a₁ or simply a

Common difference:d = aₙ − aₙ₋₁ (arithmetic only)

Common ratio:r = aₙ / aₙ₋₁ (geometric only)

Partial sum:Sₙ = a₁ + a₂ + a₃ + ... + aₙ

Sigma notation:∑(r=1 to n) f(r) = f(1) + f(2) + ... + f(n)

Infinite series:S∞ = lim(n→∞) Sₙ (exists only if series converges)

Sigma Notation — Reading and Writing

Sigma notation (∑) is a compact way to write a sum. KNEC examiners use it frequently. You must be able to expand sigma notation into its terms and write a given sum in sigma form.

Worked Examples — Sigma Notation

Ex 4 Expand: ∑(r=1 to 5) (2r + 1)

r = 12(1) + 1 = 3

r = 22(2) + 1 = 5

r = 32(3) + 1 = 7

r = 42(4) + 1 = 9

r = 52(5) + 1 = 11

Sum3 + 5 + 7 + 9 + 11 = 35

∑(r=1 to 5)(2r+1) = 35

Ex 5 Write in sigma notation: 4 + 8 + 12 + 16 + 20

PatternEach term is a multiple of 4: 4×1, 4×2, 4×3, 4×4, 4×5. General term: 4r

Limitsr runs from 1 to 5.

∑(r=1 to 5) 4r

Junior Secondary — CBC Grade 7–9

Number Patterns (CBC Grade 7–9)

Grade 7–9

In the CBC Junior Secondary curriculum (Grades 7–9), students first encounter sequences as number patterns. The focus is on identifying the rule, extending patterns, and finding the nth term. This lays the foundation for the formal AP and GP work in Senior Secondary.

Method: Finding the nth Term of a Linear Sequence

A linear (arithmetic) pattern has a constant first difference. The nth term formula is always of the form aₙ = dn + c, where d is the common difference and c is found by substituting any known term.

Find the first difference between consecutive terms — this is d.
Write the formula as aₙ = dn + c.
Substitute n = 1 (and use a₁ = first term) to find c.
Write the final formula and verify with at least two other terms.

Worked Examples — Grade 7–9 Number Patterns

Ex 6 Find the nth term of: 5, 9, 13, 17, 21, ...

Step 1First difference: 9 − 5 = 4. So d = 4.

Step 2Write: aₙ = 4n + c.

Step 3When n = 1: a₁ = 5 → 4(1) + c = 5 → c = 1.

Formulaaₙ = 4n + 1.

Verifyn = 3: 4(3) + 1 = 13 ✔ n = 5: 4(5) + 1 = 21 ✔

aₙ = 4n + 1

Ex 7 A pattern: 2, 6, 12, 20, 30, ... What is the nth term?

1st diff4, 6, 8, 10 → not constant → not linear.

2nd diff6−4 = 2, 8−6 = 2, 10−8 = 2 → constant 2nd difference → quadratic.

FormTry aₙ = An² + Bn + C.

Systemn=1: A + B + C = 2. n=2: 4A + 2B + C = 6. n=3: 9A + 3B + C = 12.

SolveEq2 − Eq1: 3A + B = 4. Eq3 − Eq2: 5A + B = 6. Subtract: 2A = 2 → A = 1. B = 4 − 3 = 1. C = 2 − 1 − 1 = 0.

Verifyn = 4: 1(16) + 1(4) + 0 = 20 ✔ n = 5: 25 + 5 = 30 ✔

aₙ = n² + n = n(n + 1)

Ex 8 Which term of the pattern 3, 7, 11, 15, ... equals 99?

nth termd = 4, a = 3. So aₙ = 3 + (n−1)(4) = 4n − 1.

Set = 994n − 1 = 99 → 4n = 100 → n = 25.

Verifya₂₅ = 4(25) − 1 = 99 ✔

The 25th term equals 99

Senior Secondary — KCSE Core

Arithmetic Progressions (AP)

Senior Secondary

An Arithmetic Progression (AP) is a sequence where each term is obtained by adding a fixed value — the common difference d — to the previous term. APs model situations with uniform increase or decrease, such as a salaried employee's annual earnings growing by a fixed increment each year.

AP Core Formulae — KCSE Examinable

nth term:aₙ = a + (n − 1)d

Sum of n terms:Sₙ = n/2 · (2a + (n−1)d)

Alternate sum:Sₙ = n/2 · (first + last)

Common difference:d = aₙ − aₙ₋₁ (always constant in an AP)

Number of terms:n = (last − first)/d + 1

Arithmetic mean:AM of a and b = (a + b)/2

Understanding the Sum Formula

The formula Sₙ = n/2(2a + (n−1)d) is derived by writing the sum forwards and backwards, then adding them term by term. Every pair of terms adds to (first term + last term), and there are n such pairs, so Sₙ = n/2 × (first + last). This is the "Gauss method" — the young Gauss famously used it to sum 1 to 100 instantly.

Worked Examples — Arithmetic Progressions

Ex 9 Find the 20th term and sum of 20 terms of: 7, 11, 15, 19, ...

Identifya = 7, d = 11 − 7 = 4, n = 20.

20th terma₂₀ = 7 + (20−1) × 4 = 7 + 76 = 83.

SumS₂₀ = 20/2 × (7 + 83) = 10 × 90 = 900.

AltS₂₀ = 20/2(2×7 + 19×4) = 10(14 + 76) = 10(90) = 900 ✔

a₂₀ = 83 | S₂₀ = 900

Ex 10 The 5th term of an AP is 23 and the 10th term is 43. Find a, d, and S₁₅.

Setupa + 4d = 23 ... (i) a + 9d = 43 ... (ii)

(ii)−(i)5d = 20 → d = 4.

Find aa = 23 − 4(4) = 23 − 16 = a = 7.

S₁₅S₁₅ = 15/2 × (2×7 + 14×4) = 7.5 × (14 + 56) = 7.5 × 70 = 525.

Verifya₅ = 7 + 4(4) = 23 ✔ a₁₀ = 7 + 9(4) = 43 ✔

a = 7, d = 4, S₁₅ = 525

Ex 11 (KCSE style) How many terms of the AP: 3, 8, 13, ... must be added to give a sum of 1005?

Identifya = 3, d = 5. Sₙ = 1005.

Applyn/2(2×3 + (n−1)×5) = 1005 → n/2(6 + 5n − 5) = 1005 → n(5n + 1)/2 = 1005.

Simplify5n² + n = 2010 → 5n² + n − 2010 = 0.

Quadraticn = [−1 ± √(1 + 4×5×2010)] / (2×5) = [−1 ± √40201] / 10 = [−1 ± 200.5...] / 10.

SolveTaking positive root: n = (−1 + 201) / 10 = 200/10 = 20. (n must be a positive integer)

VerifyS₂₀ = 20/2(6 + 19×5) = 10(6+95) = 10(101) = 1010 — wait, let me re-examine: 5(400) + 20 − 2010 = 2000 + 20 − 2010 = 10 ≠ 0. Use discriminant: n = (−1 + √40201)/10. √40201 ≈ 200.5 → n ≈ 19.95 → not integer. Try n = 20: 5(400)+20 = 2020 ≠ 2010. This AP gives no integer answer — a useful exam insight: always check if the answer is an integer!

5n² + n − 2010 = 0 → no integer solution. The sum never equals exactly 1005 for this AP.

Ex 12 Sum of all integers from 1 to 200 that are divisible by 7.

List startFirst term a = 7. Last term ≤ 200: 7 × 28 = 196. So last = 196. d = 7.

No. termsn = (196 − 7)/7 + 1 = 189/7 + 1 = 27 + 1 = 28.

SumS = 28/2 × (7 + 196) = 14 × 203 = 2842.

Sum = 2842

Ex 13 Insert 4 arithmetic means between 3 and 18.

SetupThe sequence is: 3, _, _, _, _, 18. Total 6 terms, so n = 6.

Find da₆ = a + 5d → 18 = 3 + 5d → 5d = 15 → d = 3.

Insert3 + 3 = 6, 6 + 3 = 9, 9 + 3 = 12, 12 + 3 = 15.

3, 6, 9, 12, 15, 18

Senior Secondary — KCSE Core

Geometric Progressions (GP)

Senior Secondary

A Geometric Progression (GP) is a sequence where each term is obtained by multiplying the previous term by a fixed value — the common ratio r. GPs model exponential growth and decay: population growth, compound interest, radioactive decay, and depreciation are all geometric in nature.

GP Core Formulae — KCSE Examinable

nth term:aₙ = arⁿ⁻¹

Sum (r ≠ 1):Sₙ = a(rⁿ − 1)/(r − 1) or Sₙ = a(1 − rⁿ)/(1 − r)

Infinite sum (|r| < 1):S∞ = a / (1 − r)

Common ratio:r = aₙ / aₙ₋₁ (constant throughout)

Geometric mean:GM of a and b = √(ab) (positive roots)

Condition for S∞:Series converges only when |r| < 1 (i.e. −1 < r < 1)

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Which GP sum formula to use? Use a(rⁿ−1)/(r−1) when r > 1 (easier arithmetic). Use a(1−rⁿ)/(1−r) when r < 1 (avoids negatives). Both are identical algebraically.

Worked Examples — Geometric Progressions

Ex 14 Find the 8th term and sum of first 8 terms of: 3, 6, 12, 24, ...

Identifya = 3, r = 6/3 = 2, n = 8.

8th terma₈ = 3 × 2⁷ = 3 × 128 = 384.

SumS₈ = 3(2⁸ − 1)/(2 − 1) = 3(256 − 1)/1 = 3 × 255 = 765.

a₈ = 384, S₈ = 765

Ex 15 The 3rd term of a GP is 20 and the 6th term is 160. Find a, r, and S₁₀.

Setupar² = 20 ... (i) ar⁵ = 160 ... (ii)

(ii)÷(i)r³ = 160/20 = 8 → r = 2.

Find aa(4) = 20 → a = 5.

S₁₀S₁₀ = 5(2¹⁰ − 1)/(2−1) = 5 × 1023 = 5115.

Verifya₃ = 5(4) = 20 ✔ a₆ = 5(32) = 160 ✔

a = 5, r = 2, S₁₀ = 5115

Ex 16 Find the infinite sum of: 8 + 4 + 2 + 1 + 0.5 + ...

Identifya = 8, r = 4/8 = 1/2. Since |r| = 1/2 < 1, the infinite sum exists.

S∞S∞ = 8 / (1 − 1/2) = 8 / (1/2) = 8 × 2 = 16.

S∞ = 16

Ex 17 Prove that 0.̄3 = 1/3 using geometric series.

Write0.333... = 3/10 + 3/100 + 3/1000 + ...

GP forma = 3/10, r = 1/10. |r| = 0.1 < 1, so infinite sum exists.

ApplyS∞ = (3/10) / (1 − 1/10) = (3/10) / (9/10) = (3/10) × (10/9) = 3/9 = 1/3.

0.̄3 = 1/3 ✓ (QED)

Ex 18 (KCSE Application) A car depreciates at 15% per year. Original value: Ksh 800,000. Find value after 5 years and the total depreciation over 5 years.

SetupEach year value = 85% of previous → r = 0.85. a = 800,000.

After 5 yra₆ = 800,000 × (0.85)⁵ = 800,000 × 0.4437... ≈ Ksh 355,014.

SumS₅ = 800,000(1 − 0.85⁵)/(1 − 0.85) = 800,000 × (1 − 0.4437)/0.15 = 800,000 × 3.709 ≈ Ksh 2,967,200.

DepreciationTotal depreciation = 5 × 800,000 − S₅... Note: the sum S₅ = total book values, not a depreciation sum. Value dropped from 800,000 to 355,014 → total loss ≈ Ksh 444,986.

Value after 5 years ≈ Ksh 355,014. Total value lost ≈ Ksh 444,986.

Ex 19 Insert 3 geometric means between 2 and 162.

SetupSequence: 2, _, _, _, 162. Total 5 terms. a = 2, a₅ = 162.

Find rar⁴ = 162 → 2r⁴ = 162 → r⁴ = 81 → r = 3 (taking positive root).

Insert2×3 = 6, 6×3 = 18, 18×3 = 54.

2, 6, 18, 54, 162

A-Level / Senior Secondary

Special Summation Formulas

A-Level

Beyond AP and GP, there are closed-form formulas for summing the first n natural numbers, their squares, and their cubes. These are essential for KNEC Advanced Level examinations and university entrance tests. Mastering them allows evaluation of sums involving polynomials in r.

Standard Summation Results

∑ 1 (r=1 to n):= n

∑ r (r=1 to n):= n(n+1)/2

∑ r² (r=1 to n):= n(n+1)(2n+1)/6

∑ r³ (r=1 to n):= [n(n+1)/2]² = (∑r)²

Linearity property:∑[af(r) + bg(r)] = a·∑f(r) + b·∑g(r)

Worked Examples — Special Summation Formulas

Ex 20 Evaluate ∑(r=1 to 12) r²

Formula∑r² = n(n+1)(2n+1)/6. Here n = 12.

Compute= 12 × 13 × 25 / 6 = (12/6) × 13 × 25 = 2 × 13 × 25 = 650.

∑(r=1 to 12) r² = 650

Ex 21 Find ∑(r=1 to n)(3r² − 2r + 5) in terms of n.

Split= 3∑r² − 2∑r + 5∑1

Substitute= 3·[n(n+1)(2n+1)/6] − 2·[n(n+1)/2] + 5n

Simplify= n(n+1)(2n+1)/2 − n(n+1) + 5n

Factor n= n [ (n+1)(2n+1)/2 − (n+1) + 5 ]

Expand= n [ (2n²+3n+1)/2 − n − 1 + 5 ]

Combine= n [ (2n²+3n+1 − 2n − 2 + 10)/2 ] = n(2n²+n+9)/2

∑(3r²−2r+5) = n(2n²+n+9)/2

Ex 22 Evaluate ∑(r=5 to 15) r³

Strategy∑(r=5 to 15) r³ = ∑(r=1 to 15) r³ − ∑(r=1 to 4) r³

n=15= [15×16/2]² = 120² = 14400

n=4= [4×5/2]² = 10² = 100

Answer= 14400 − 100 = 14300.

∑(r=5 to 15) r³ = 14300

A-Level / Senior Secondary

Telescoping Series

A-Level

A telescoping series is one where most terms cancel with adjacent terms, leaving only a small number of terms in the final sum — just as a telescope collapses. They are common in KNEC advanced examinations and require expressing the general term as a difference of two functions: f(r) − f(r+1).

Telescoping Principle

∑(r=1 to n) [f(r) − f(r+1)] = f(1) − f(n+1)

The key technique is partial fractions. When a general term has a product of linear factors in the denominator, use partial fractions to express it as a difference, then apply the telescoping principle.

Worked Examples — Telescoping Series

Ex 23 Find ∑(r=1 to n) 1/(r(r+1))

P. Fractions1/(r(r+1)) = 1/r − 1/(r+1). Verify: (r+1−r)/(r(r+1)) = 1/(r(r+1)) ✔

Write sum∑ [1/r − 1/(r+1)] = (1/1 − 1/2) + (1/2 − 1/3) + (1/3 − 1/4) + ... + (1/n − 1/(n+1))

TelescopeAll middle terms cancel. Remaining: 1 − 1/(n+1) = n/(n+1).

∑(r=1 to n) 1/(r(r+1)) = n/(n+1)

Ex 24 Find ∑(r=1 to 20) 1/((2r−1)(2r+1))

P. Fractions1/((2r−1)(2r+1)) = (1/2)[1/(2r−1) − 1/(2r+1)]

Sum(1/2) × ∑(r=1 to 20)[1/(2r−1) − 1/(2r+1)]

Telescope= (1/2)[1/1 − 1/41] = (1/2)[40/41] = 20/41.

∑ 1/((2r−1)(2r+1)) = 20/41

A-Level / KNEC Advanced

Mixed & Combined Series

A-Level

Some KNEC examination questions combine an AP and a GP in the same series — for example, an arithmetico-geometric series where each term is a product of an AP term and a GP term. These require the difference method or the standard AP–GP product formula.

Arithmetico-Geometric Series Sum

S = a/(1−r) + dr/(1−r)² (for |r| < 1, infinite series) S_n = [a − (a+(n−1)d)rⁿ]/(1−r) + dr(1−rⁿ)/(1−r)² (finite)

Worked Example — Arithmetico-Geometric Series

Ex 25 Find the sum: 1 + 2×(1/2) + 3×(1/4) + 4×(1/8) + ... (infinite)

IdentifyGeneral term: n × (1/2)ⁿ⁻¹. AP part: a=1, d=1. GP part: r=1/2.

Let SS = 1 + 2(1/2) + 3(1/4) + 4(1/8) + ...

×(1/2)S/2 = 1(1/2) + 2(1/4) + 3(1/8) + 4(1/16) + ...

SubtractS − S/2 = 1 + (1/2) + (1/4) + (1/8) + ... (infinite GP)

S/2= 1/(1 − 1/2) = 2.

AnswerS = 2 × 2 = 4.

S = 4

A-Level / Senior Secondary

Recurrence Relations

A-Level

A recurrence relation (also called a recursive formula) defines each term of a sequence in terms of one or more previous terms. Unlike explicit formulas, recurrence relations require you to know earlier terms before computing later ones. Fibonacci sequences and many financial models use recurrence relations.

Worked Examples — Recurrence Relations

Ex 26 A sequence is defined by uₙ₊₁ = 3uₙ − 2, with u₁ = 4. Find u₂, u₃, u₄ and describe the type of sequence.

u₂= 3(4) − 2 = 12 − 2 = 10

u₃= 3(10) − 2 = 28

u₄= 3(28) − 2 = 82

TypeDifferences: 6, 18, 54 → each multiplied by 3. This is a geometric-like sequence (though shifted by a constant). The differences form a GP.

u₂ = 10, u₃ = 28, u₄ = 82

Ex 27 The Fibonacci sequence is defined by F₁ = 1, F₂ = 1, Fₙ = Fₙ₋₁ + Fₙ₋₂. Write the first 10 terms and find the ratio Fₙ/Fₙ₋₁ as n grows large.

Terms1, 1, 2, 3, 5, 8, 13, 21, 34, 55

Ratios1/1=1, 2/1=2, 3/2=1.5, 5/3=1.667, 8/5=1.6, 13/8=1.625, 21/13≈1.615, 34/21≈1.619, 55/34≈1.618

LimitThe ratio converges to φ = (1+√5)/2 ≈ 1.618 — the Golden Ratio.

First 10: 1,1,2,3,5,8,13,21,34,55. Ratio → φ ≈ 1.618

University Level

Convergence Tests

University

An infinite series converges if its partial sums approach a finite limit. Determining whether a series converges or diverges — and if so, to what value — is a central topic in university-level analysis. The CBC Senior Secondary curriculum introduces convergence conceptually; full formal tests are examined at university and A-level.

Test

Condition for Convergence / Divergence

nth-term test

If lim(aₙ) ≠ 0 → diverges. (If lim = 0, test is inconclusive)

Geometric series

Converges iff |r| < 1 with sum a/(1−r)

p-series ∑1/nᵖ

Converges iff p > 1. Diverges if p ≤ 1

Ratio test (D'Alembert)

L = lim|aₙ₊₁/aₙ|. L<1 → converges; L>1 → diverges; L=1 → inconclusive

Root test (Cauchy)

L = lim|aₙ|^(1/n). Same rules as ratio test

Integral test

If f is positive, continuous, decreasing: ∑f(n) and ∫f(x)dx converge or diverge together

Comparison test

0 ≤ aₙ ≤ bₙ: if ∑bₙ converges → ∑aₙ converges; if ∑aₙ diverges → ∑bₙ diverges

Limit comparison

If lim(aₙ/bₙ) = L (finite, L>0): ∑aₙ and ∑bₙ both converge or both diverge

Alternating (Leibniz)

∑(−1)ⁿaₙ converges if aₙ is decreasing → 0

Worked Examples — Convergence Tests

Ex 28 Test for convergence: ∑(n=1 to ∞) n!/nⁿ

TestUse the Ratio test. aₙ = n!/nⁿ.

aₙ₊₁= (n+1)!/(n+1)ⁿ⁺¹

Ratio|aₙ₊₁/aₙ| = [(n+1)! × nⁿ] / [(n+1)ⁿ⁺¹ × n!] = (n+1) × nⁿ / (n+1)ⁿ⁺¹ = nⁿ/(n+1)ⁿ = (n/(n+1))ⁿ = (1/(1+1/n))ⁿ

Limitlim(n→∞) (1/(1+1/n))ⁿ = 1/e ≈ 0.368 < 1.

ResultSince L = 1/e < 1, the series converges by the Ratio test.

Converges (L = 1/e < 1)

Ex 29 Does ∑(n=1 to ∞) 1/n² converge?

IdentifyThis is a p-series with p = 2 > 1.

ResultBy the p-series test, it converges. (Its sum is π²/6 ≈ 1.645 — Basel problem)

Converges (p = 2 > 1). Sum = π²/6.

Ex 30 Does ∑(n=1 to ∞) 1/n converge? (The harmonic series)

p-seriesp = 1. Since p ≤ 1, the p-series test says it diverges.

NoteThe harmonic series diverges even though individual terms → 0. This is a crucial exam example: lim(aₙ) = 0 is necessary but NOT sufficient for convergence.

Diverges (p = 1). Classic counterexample to "terms → 0 implies convergence".

University Level

Power Series & Radius of Convergence

University

A power series is an infinite series whose terms involve powers of a variable x. They are the building blocks of modern analysis and are used to define functions, solve differential equations, and compute values of transcendental functions to any desired accuracy.

Power Series — Key Concepts

General form:∑(n=0 to ∞) aₙ(x−c)ⁿ = a₀ + a₁(x−c) + a₂(x−c)² + ...

Radius of conv. R:R = lim(n→∞) |aₙ/aₙ₊₁| (ratio of consecutive coefficients)

Interval of conv.:(c − R, c + R) — always test endpoints separately

R = 0:Series converges only at x = c

R = ∞:Series converges for all x ∈ ℝ

Worked Examples — Power Series

Ex 31 Find the radius of convergence of ∑(n=0 to ∞) xⁿ/n!

Ratio testaₙ = xⁿ/n! → |aₙ₊₁/aₙ| = |x|ⁿ⁺¹/(n+1)! × n!/|x|ⁿ = |x|/(n+1)

Limitlim(n→∞) |x|/(n+1) = 0 < 1 for all finite x.

ResultR = ∞. The series converges for all x ∈ ℝ. (This is the power series for eˣ.)

R = ∞ (converges everywhere)

Ex 32 Find the interval of convergence of ∑(n=1 to ∞) xⁿ/n

Ratio test|aₙ₊₁/aₙ| = |x|ⁿ⁺¹/(n+1) × n/|x|ⁿ = n|x|/(n+1) → |x| as n → ∞.

Convergeswhen |x| < 1 → −1 < x < 1. R = 1.

Endpointsx = 1: ∑1/n (harmonic series) → diverges. x = −1: ∑(−1)ⁿ/n (alternating harmonic) → converges.

Interval−1 ≤ x < 1. (This is the series for −ln(1−x).)

Interval of convergence: [−1, 1)

University Level

Maclaurin & Taylor Series

University

A Maclaurin series is a power series expansion of a function f(x) about x = 0. A Taylor series generalises this to any centre point x = a. They allow us to approximate complex functions (like sin x, eˣ, ln x) using polynomials — the foundation of numerical methods used in engineering, physics, and computer science.

Standard Maclaurin Series — Memorise for KNEC

eˣ:1 + x + x²/2! + x³/3! + x⁴/4! + ... (R = ∞)

sin x:x − x³/3! + x⁵/5! − x⁷/7! + ... (R = ∞)

cos x:1 − x²/2! + x⁴/4! − x⁶/6! + ... (R = ∞)

ln(1+x):x − x²/2 + x³/3 − x⁴/4 + ... (R = 1, −1 < x ≤ 1)

(1+x)ⁿ:1 + nx + n(n−1)x²/2! + n(n−1)(n−2)x³/3! + ... (R = 1)

1/(1−x):1 + x + x² + x³ + ... (R = 1, |x| < 1) — geometric series

General Maclaurin / Taylor Formulas

Maclaurin: f(x) = f(0) + f'(0)x + f''(0)x²/2! + f'''(0)x³/3! + ... Taylor: f(x) = f(a) + f'(a)(x−a) + f''(a)(x−a)²/2! + ...

Worked Examples — Maclaurin & Taylor Series

Ex 33 Find the Maclaurin series for f(x) = eˣ up to the x⁴ term, and use it to estimate e⁰·² to 4 d.p.

Derivativesf(x) = f'(x) = f''(x) = ... = eˣ. At x = 0: all derivatives = 1.

Serieseˣ = 1 + x + x²/2 + x³/6 + x⁴/24 + ...

x = 0.2e⁰·² ≈ 1 + 0.2 + 0.04/2 + 0.008/6 + 0.0016/24

Compute= 1 + 0.2 + 0.02 + 0.001333... + 0.0000667 ≈ 1.2214

Actuale⁰·² = 1.22140... ✔ (accurate to 4 d.p.)

eˣ = 1 + x + x²/2! + x³/3! + ...; e⁰·² ≈ 1.2214

Ex 34 Using the Maclaurin series for sin x, evaluate sin(0.5 rad) correct to 5 d.p.

Seriessin x = x − x³/6 + x⁵/120 − x⁷/5040 + ...

x = 0.5sin(0.5) ≈ 0.5 − (0.125)/6 + (0.03125)/120 − (0.0078125)/5040

Compute= 0.5 − 0.020833 + 0.000260 − 0.0000016 ≈ 0.47943

Actualsin(0.5) = 0.47943... ✔

sin(0.5) ≈ 0.47943

Ex 35 Derive the first three non-zero terms of the Maclaurin series for ln(1+x) from first principles.

f(x)f(x) = ln(1+x) → f(0) = 0

f'(x)= 1/(1+x) → f'(0) = 1

f''(x)= −1/(1+x)² → f''(0) = −1

f'''(x)= 2/(1+x)³ → f'''(0) = 2

Applyln(1+x) = 0 + 1·x + (−1)x²/2! + 2·x³/3! + ... = x − x²/2 + x³/3 − ...

ln(1+x) = x − x²/2 + x³/3 − ... (|x| ≤ 1, x ≠ −1)

Quick Reference

Complete Formula Sheet

All Levels

All key formulas for Sequences & Series in one place. Print or screenshot this section as a revision aid for KCSE and university assessments.

Arithmetic Progression

aₙ = a + (n−1)d

Sₙ = n/2(2a + (n−1)d)

Sₙ = n/2(first + last)

d = aₙ − aₙ₋₁

n = (L − a)/d + 1

Geometric Progression

aₙ = arⁿ⁻¹

Sₙ = a(rⁿ−1)/(r−1), r≠1

S∞ = a/(1−r), |r|<1

r = aₙ/aₙ₋₁

GM of a,b = √(ab)

Summation Formulas

∑1 = n

∑r = n(n+1)/2

∑r² = n(n+1)(2n+1)/6

∑r³ = [n(n+1)/2]²

Telescoping & Partial Fractions

∑[f(r)−f(r+1)] = f(1)−f(n+1)

1/(r(r+1)) = 1/r − 1/(r+1)

∑1/(r(r+1)) = n/(n+1)

Maclaurin Series

eˣ = ∑xⁿ/n! (R=∞)

sin x = ∑(−1)ⁿx²ⁿ⁺¹/(2n+1)!

cos x = ∑(−1)ⁿx²ⁿ/(2n)!

ln(1+x) = ∑(−1)ⁿ⁺¹xⁿ/n

(1+x)ⁿ = ∑C(n,r)xʳ

Convergence Tests

Ratio: L = lim|aₙ₊₁/aₙ|

Root: L = lim|aₙ|^(1/n)

p-series: conv. iff p>1

Geometric: conv. iff |r|<1

Alternating: conv. if aₙ↘0

Power Series

∑aₙ(x−c)ⁿ centred at c

R = lim|aₙ/aₙ₊₁|

Interval: (c−R, c+R)

Check endpoints separately

Special Means

AM of a,b = (a+b)/2

GM of a,b = √(ab)

AM ≥ GM ≥ HM (AM-GM inequality)

HM of a,b = 2ab/(a+b)

Exam Guidance

Common Errors to Avoid

All Levels

These are the most frequent mistakes Kenyan students make in Sequences & Series — drawn from analysis of past KCSE marking schemes and common error patterns at university level.

✗ WrongFor GP: Using Sₙ = n/2(2a+(n−1)d) instead of GP formula

✓ RightSₙ = a(rⁿ−1)/(r−1) for GP. The AP sum formula only applies to APs.

Always identify the type FIRST before writing any formula. Check: is the difference constant (AP) or is the ratio constant (GP)?

✗ WrongAssuming S∞ exists for any GP, e.g. claiming S∞ for r = 2

✓ RightS∞ = a/(1−r) only exists when |r| < 1. If |r| ≥ 1, the infinite sum diverges.

Always state and check the condition |r| < 1 before using the infinite GP sum formula.

✗ WrongWriting aₙ = arⁿ instead of arⁿ⁻¹ (off by one error)

✓ Rightaₙ = arⁿ⁻¹. The first term (n=1) gives a₁ = ar⁰ = a, not ar.

Always verify your formula by substituting n = 1 to confirm you get the first term a.

✗ WrongUsing ∑r² = [n(n+1)/2]² (confusing with ∑r³)

✓ Right∑r² = n(n+1)(2n+1)/6. It is ∑r³ = [n(n+1)/2]².

Memorise the distinction: ∑r³ equals (∑r)², but ∑r² has the (2n+1) factor in the numerator.

✗ WrongConcluding a series converges because its terms approach zero

✓ Rightlim aₙ = 0 is necessary but NOT sufficient. The harmonic series ∑1/n diverges even though 1/n → 0.

Apply a proper convergence test (ratio test, p-series, comparison) to confirm convergence.

✗ WrongForgetting to check endpoints when finding interval of convergence

✓ RightThe ratio test only gives the open interval (c−R, c+R). Each endpoint must be tested separately with a different test.

An endpoint might converge (include it with [) or diverge (exclude it with ()). Never assume without checking.

✗ WrongMisidentifying a quadratic sequence as arithmetic

✓ RightFind SECOND differences. If second differences are constant (not first), the sequence is quadratic, not arithmetic.

Grade 7–9 CBC sequences: always check first AND second differences before concluding the type.

✗ WrongIn finding the number of terms, giving a non-integer answer and ignoring it

✓ RightThe number of terms n must be a positive integer. If your equation gives n = 7.3, re-check the problem — either your working has an error or no exact integer solution exists.

This is a critical checking step that earns marks in KCSE for showing verification.