Sequences and Proof

This lesson covers proving properties of sequences, including convergence and divergence, and introduces proof by induction as awareness content for the Edexcel A-Level (9MA0) specification. Proof is a key mathematical skill and sequence-based proofs appear frequently in exams.

---

Types of Proof Relevant to Sequences

Proof type	When to use
Direct proof	Showing a statement is true using logical steps from known results
Proof by contradiction	Assuming the opposite and reaching a contradiction
Proof by exhaustion	Checking all possible cases
Proof by induction	Proving a statement for all positive integers (A-Level awareness)

---

Proving Sequence Properties Directly

Worked Example 1

Prove that the sum of the first n odd numbers is n².

The first n odd numbers are: 1, 3, 5, 7, ..., (2n - 1).

This is an AP with a = 1, d = 2, and the nth term is 2n - 1.

S(n) = n/2 x (first + last) = n/2 x (1 + (2n - 1)) = n/2 x 2n = n²

Worked Example 2

Prove that if three consecutive terms of an AP are a - d, a, and a + d, then their sum is always 3a.

Sum = (a - d) + a + (a + d) = 3a

The d terms cancel, confirming the sum depends only on the middle term.

Worked Example 3

Prove that the product of any three consecutive integers is divisible by 6.

Let the three consecutive integers be n, n + 1, n + 2.

Among any three consecutive integers:

• At least one is divisible by 2 (consecutive integers alternate even/odd)
• At least one is divisible by 3 (every third integer is a multiple of 3)

Since 2 and 3 are coprime, the product is divisible by 2 x 3 = 6.

---

Proving Convergence

Worked Example 4

A sequence is defined by u(n+1) = (u(n)² + 6) / (2u(n)), u(1) = 3.

(a) Show that if the sequence converges to a limit L, then L = sqrt(6).

If the sequence converges, u(n) tends to L and u(n+1) tends to L:

L = (L² + 6) / (2L)

2L² = L² + 6

L² = 6

L = sqrt(6) (taking the positive root since all terms are positive)

(b) Compute the first few terms to verify convergence.

u(1) = 3 u(2) = (9 + 6) / 6 = 15/6 = 2.5 u(3) = (6.25 + 6) / 5 = 12.25/5 = 2.45 u(4) = (6.0025 + 6) / 4.9 = 12.0025/4.9 ≈ 2.44949...

sqrt(6) ≈ 2.44949..., confirming rapid convergence.

---

Proving Divergence

Worked Example 5

Show that the sequence u(n) = (-1)^n x n diverges.

The terms are: -1, 2, -3, 4, -5, 6, ...

For even n: u(n) = n, which tends to infinity For odd n: u(n) = -n, which tends to negative infinity

Since the terms do not approach any single value, the sequence diverges.

Worked Example 6

Prove that the geometric sequence with first term 3 and common ratio 2 diverges.

u(n) = 3 x 2^(n-1)

As n tends to infinity, 2^(n-1) tends to infinity, so u(n) tends to infinity.

Since the terms grow without bound, the sequence diverges.

---

Proof by Induction (Awareness)

Proof by induction is a powerful technique for proving statements about positive integers. At A-Level, you should understand the principle and be able to follow a proof.

The Structure of Induction

1. Base case: Verify the statement for n = 1 (or the smallest relevant n).
2. Inductive hypothesis: Assume the statement is true for n = k.
3. Inductive step: Prove the statement is true for n = k + 1, using the assumption from step 2.
4. Conclusion: By induction, the statement is true for all positive integers n.

Worked Example 7

Prove by induction that ∑ (r = 1 to n) r = n(n + 1)/2.

Base case (n = 1): LHS = 1. RHS = 1(2)/2 = 1. LHS = RHS.

Inductive hypothesis: Assume ∑ (r = 1 to k) r = k(k + 1)/2 for some positive integer k.

Inductive step: Consider n = k + 1.

∑ (r = 1 to k+1) r = [∑ (r = 1 to k) r] + (k + 1)

= k(k + 1)/2 + (k + 1) [using the hypothesis]

= k(k + 1)/2 + 2(k + 1)/2

= (k + 1)(k + 2)/2

This is the formula with n = k + 1.

Conclusion: The statement is true for n = 1, and if true for n = k then true for n = k + 1. Therefore, by mathematical induction, ∑ (r = 1 to n) r = n(n + 1)/2 for all positive integers n.

Worked Example 8

Prove by induction that u(n) = 3^n - 1 where u(1) = 2 and u(n+1) = 3u(n) + 2.

Base case (n = 1): u(1) = 3^1 - 1 = 2. Given u(1) = 2.

Inductive hypothesis: Assume u(k) = 3^k - 1 for some positive integer k.

Inductive step:

u(k + 1) = 3u(k) + 2

= 3(3^k - 1) + 2 [using hypothesis]

= 3^(k+1) - 3 + 2

= 3^(k+1) - 1

This matches the formula with n = k + 1.

Conclusion: By mathematical induction, u(n) = 3^n - 1 for all positive integers n.

---

Proving Sequence Identities

Worked Example 9

Prove that for a GP, S(2n)/S(n) = 1 + r^n.

S(2n) = a(1 - r^(2n)) / (1 - r) S(n) = a(1 - r^n) / (1 - r)

S(2n)/S(n) = (1 - r^(2n)) / (1 - r^n)

= (1 - r^n)(1 + r^n) / (1 - r^n)

= 1 + r^n

The ratio is independent of a.

Worked Example 10

Prove that the sum of the first n terms of 1/(1x2) + 1/(2x3) + 1/(3x4) + ... is n/(n+1).

Each term can be written using partial fractions:

1/(r(r+1)) = 1/r - 1/(r+1)

So ∑ (r = 1 to n) 1/(r(r+1)) = ∑ (r = 1 to n) [1/r - 1/(r+1)]

This is a telescoping sum: = (1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + ... + (1/n - 1/(n+1))

= 1 - 1/(n+1)

= n/(n+1)

---

Exam Tips

Tip	Detail
Proof layout	Use clear headings: "Base case", "Inductive hypothesis", "Inductive step", "Conclusion"
Convergence proofs	Set u(n+1) = u(n) = L, solve, and reject impossible roots
Telescoping	Look for partial fractions when terms cancel in pairs
Be rigorous	State what you assume, what you need to prove, and why the conclusion follows
Divergence	Show terms grow without bound OR oscillate without settling

---

Practice Problems

1. Prove that the sum of the first n even numbers is n(n + 1). [Hint: the kth even number is 2k]
2. Prove by induction that ∑ (r = 1 to n) r² = n(n + 1)(2n + 1)/6.
3. A sequence is defined by u(n+1) = sqrt(2u(n) + 3), u(1) = 1. Find the limit L. [Answer: L = 3]
4. Show that ∑ (r = 1 to n) 1/((2r-1)(2r+1)) = n/(2n+1). [Hint: partial fractions]
5. Prove that for a GP with |r| < 1, S(infinity) - S(n) = ar^n / (1 - r).

---

Summary

• Direct proof is the most common method for sequence identities.
• To prove convergence, set u(n+1) = u(n) = L and solve; reject values inconsistent with the sequence.
• Proof by induction has four parts: base case, hypothesis, inductive step, conclusion.
• Telescoping sums arise when terms cancel in pairs (often via partial fractions).
• Always present proofs clearly and logically — marks are awarded for structure as well as correctness.

A-Level Deep Dive: Sequences and Proof

Spec mapping

Edexcel 9MA0-01 specification section 4 — Sequences and series, sub-strand 4.4 covers sequences and series in modelling and proof of standard results (refer to the official specification document for exact wording). On the 9MA0 specification this refers explicitly to direct, algebraic proofs of formulae such as the sum of an arithmetic progression $S_n = \tfrac{n}{2}(2a + (n-1)d)$Sn​=2n​(2a+(n−1)d) and the sum of a geometric progression $S_n = \dfrac{a(1 - r^n)}{1 - r}$Sn​=1−ra(1−rn)​ for $r \neq 1$r=1. Proof by induction is not on the 9MA0 specification — it lives on the Further Mathematics 9FM0 spec (Core Pure 1, section 1). The A-Level paper will only ever ask you to derive a sequence/series identity by direct algebraic manipulation, the standard pairing trick, or a "show that" telescoping-style argument. Synoptic ties: sigma notation (\sum\); arithmetic and geometric progressions; AO2 proof-writing as a transferable skill across Pure; algebraic manipulation of polynomial expressions; combinatorial identities such as $\binom{n}{r} + \binom{n}{r+1} = \binom{n+1}{r+1}$(rn​)+(r+1n​)=(r+1n+1​).

---

Worked example with full mark scheme

Question (8 marks): A geometric series has first term $a$a and common ratio $r$r, where $r \neq 1$r=1. Prove from first principles that the sum of the first $n$n terms is

$S_n = \dfrac{a(1 - r^n)}{1 - r}.$Sn​=1−ra(1−rn)​.

Solution with mark scheme:

Step 1 — write down $S_n$Sn​ explicitly.

By definition the first $n$n terms of the GP are $a, ar, ar^2, \dots, ar^{n-1}$a,ar,ar2,…,arn−1, so

$S_n = a + ar + ar^2 + \dots + ar^{n-1}. \quad (\star)$Sn​=a+ar+ar2+⋯+arn−1.(⋆)

M1 — correct expression for $S_n$Sn​ written out term by term, with the final term $ar^{n-1}$arn−1 (not $ar^n$arn). Common error: writing the last term as $ar^n$arn, which corresponds to $n + 1$n+1 terms and produces the wrong formula.

A1 — fully correct expression including ellipsis and final term.

Step 2 — multiply both sides of $(\star)$(⋆) by $r$r.

$rS_n = ar + ar^2 + ar^3 + \dots + ar^{n-1} + ar^n. \quad (\star\star)$rSn​=ar+ar2+ar3+⋯+arn−1+arn.(⋆⋆)

M1 — correct multiplication; every index has shifted up by one. Examiners specifically check that the indices in $(\star\star)$(⋆⋆) match $(\star)$(⋆) shifted by one place.

Step 3 — subtract $(\star\star)$(⋆⋆) from $(\star)$(⋆).

Lining the two equations up so that matching powers of $r$r stack vertically, almost every term cancels:

$S_n - rS_n = a - ar^n.$Sn​−rSn​=a−arn.

M1 — recognising and executing the cancellation. The pairs $ar - ar$ar−ar, $ar^2 - ar^2$ar2−ar2, ..., $ar^{n-1} - ar^{n-1}$arn−1−arn−1 all vanish, leaving only the leading $a$a from $(\star)$(⋆) and the trailing $-ar^n$−arn from $(\star\star)$(⋆⋆).

A1 — correct difference $S_n - rS_n = a - ar^n$Sn​−rSn​=a−arn.

Step 4 — factorise both sides.

$S_n(1 - r) = a(1 - r^n).$Sn​(1−r)=a(1−rn).

M1 — clean factorisation on each side. The LHS factor $(1 - r)$(1−r) is the engine of the whole proof.

Step 5 — divide by $(1 - r)$(1−r), justifying the division.

Because we are told $r \neq 1$r=1, we have $1 - r \neq 0$1−r=0, so we can divide:

$S_n = \dfrac{a(1 - r^n)}{1 - r}.$Sn​=1−ra(1−rn)​.

A1 — final printed result with the explicit acknowledgement that $r \neq 1$r=1 is the condition that makes the division valid. Omitting this justification typically loses the final accuracy mark in proof questions.

Total: 8 marks (M4 A4).

---

Specimen question modelled on the Edexcel 9MA0 Paper 1 format

Question (6 marks): A sequence is defined for $n \geq 1$n≥1 by $u_n = 3n^2 + 2n - 1$un​=3n2+2n−1.

(a) Show that, for all $n \geq 1$n≥1, $u_{n+1} - u_n = 6n + 5$un+1​−un​=6n+5. (3)

(b) Hence prove that the differences $u_{n+1} - u_n$un+1​−un​ form an arithmetic progression, stating its first term and common difference. (3)

Mark scheme decomposition by AO:

(a)

• M1 (AO1.1a) — substituting $n + 1$n+1 correctly: $u_{n+1} = 3(n+1)^2 + 2(n+1) - 1$un+1​=3(n+1)2+2(n+1)−1, expanded as $3(n^2 + 2n + 1) + 2n + 2 - 1 = 3n^2 + 6n + 3 + 2n + 1 = 3n^2 + 8n + 4$3(n2+2n+1)+2n+2−1=3n2+6n+3+2n+1=3n2+8n+4.
• M1 (AO1.1b) — forming and simplifying the difference: $u_{n+1} - u_n = (3n^2 + 8n + 4) - (3n^2 + 2n - 1) = 6n + 5$un+1​−un​=(3n2+8n+4)−(3n2+2n−1)=6n+5.
• A1 (AO2.1) — printed result $6n + 5$6n+5 obtained, with the cancellation of the $3n^2$3n2 terms shown.

(b)

• M1 (AO2.2a) — recognising that to prove an AP one must show consecutive differences are constant, so consider $(6(n+1) + 5) - (6n + 5)$(6(n+1)+5)−(6n+5).
• A1 (AO2.1) — evaluating: $(6n + 11) - (6n + 5) = 6$(6n+11)−(6n+5)=6, a constant independent of $n$n, so the differences form an AP.
• A1 (AO2.4) — first term ($n = 1$n=1): $6(1) + 5 = 11$6(1)+5=11; common difference: $6$6. State both and conclude with a one-line "hence the differences form an AP with first term 11 and common difference 6".

Total: 6 marks split AO1 = 2, AO2 = 4. This is an AO2-heavy specimen — typical of proof questions on 9MA0, where the manipulation is routine but the marks reward precise reasoning and explicit justification of each logical step.

---

Synoptic links

Connects to:

1. Sigma notation (4.1): every sum of the form $\sum_{r=1}^{n} f(r)$∑r=1n​f(r) that appears in proof questions can be unpacked into $f(1) + f(2) + \dots + f(n)$f(1)+f(2)+⋯+f(n). The proof of $\sum_{r=1}^{n} r = \tfrac{n(n+1)}{2}$∑r=1n​r=2n(n+1)​ uses Gauss's pairing trick — write the sum forwards and backwards, add term by term to get $n$n pairs each summing to $n+1$n+1.

2. AP and GP formulae (4.2, 4.3): the proofs of $S_n = \tfrac{n}{2}(2a + (n-1)d)$Sn​=2n​(2a+(n−1)d) and $S_n = \tfrac{a(1 - r^n)}{1 - r}$Sn​=1−ra(1−rn)​ are the canonical examples on this specification. Both are derivable from first principles using the pairing/multiplication tricks and are fair game for an 8-mark "prove from first principles" question.

3. Algebraic manipulation (2.x): every proof rests on confidence with expanding brackets, factorising, and combining fractions. A neat factor of $(1 - r)$(1−r) at the right moment is what turns a four-line algebraic mess into a printed result.

4. Proof by induction (Further Maths 9FM0 Core Pure 1): induction is the natural extension method for sequence/series identities and is a standard A* enrichment topic. It is not on 9MA0 — flag it in your study notes as Further Maths content and treat any informal induction in 9MA0 solutions as outside the spec.

5. Combinatorial proofs of binomial identities (4.x and into 9MA0-02): identities such as $\binom{n}{r} + \binom{n}{r+1} = \binom{n+1}{r+1}$(rn​)+(r+1n​)=(r+1n+1​) (Pascal's rule) admit both algebraic and combinatorial proofs, mirroring the same "two valid routes" structure that appears in sequence proofs.

---

Mark-scheme literacy

Sequence-proof questions on 9MA0 split AO marks heavily toward AO2:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	25–35%	Substituting, expanding, factorising, applying sum formulae
AO2 (reasoning / interpretation)	55–70%	Justifying each logical step, stating conditions ($r \neq 1$r=1, $n \geq 1$n≥1), writing a one-line concluding statement, presenting the proof in a clear structure
AO3 (problem-solving)	0–10%	Spotting the pairing/multiplication trick when the question is open-ended

A subtle but examiner-rewarded distinction: "prove" vs "show that". "Prove" demands a fully general argument with explicit justification of each step and a closing statement. "Show that" is a slightly weaker command — examiners will accept a clean derivation with the key step demonstrated, without insisting on the same level of meta-commentary. Both, however, demand the printed answer to be reached and labelled (e.g. "as required" or "QED" or simply boxing the result).