Sequences and Series Notes (JC1 H2 Math) — Free PDF

Q: How do you find the nth term from the sum of the first n terms?

For n ≥ 2, use uₙ = Sₙ − Sₙ₋₁. Replace n by (n − 1) in Sₙ, putting brackets around every (n − 1). Then find u₁ = S₁ separately and check that it fits: for Sₙ = 3n² + 2n, uₙ = 6n − 1 and u₁ = S₁ = 5 agrees, so the formula holds for all n ≥ 1.

Q: When does a geometric series have a sum to infinity?

Only when |r| 1 the series diverges and S∞ does not exist. The condition is on |r|, so a negative ratio such as r = −2 ÷ 3 still gives a convergent series.

A sequence is a list of numbers in a definite order. A series is the sum of its terms. Everything in this topic rests on these two ideas and on the link between the nth term uₙ and Sₙ, the sum of the first n terms.

H2 Mathematics (9758) Topic 2.1 appears in almost every Promotional and A-level Paper 1. It is usually set as an AP/GP modelling question plus a method-of-differences “show that … hence” question.

What these notes cover

•Sequences and series
•Finite and infinite series
•The link between uₙ and Sₙ
•Sum of a block of terms
•Recurrence relations and the GC
•Describing the behaviour of a sequence
•Limit of a convergent recurrence
•Arithmetic progressions
•Geometric progressions and sum to infinity
•Sigma notation and change of index
•Method of differences
•Common mistakes and exam technique

Key definitions

Sequence: A list of numbers written in a definite order, where each number is a term. It is a function uₙ = f(n) with n ∈ ℤ⁺, so its graph is a set of separate points, not a continuous curve.
Series: The sum of the terms of a sequence. Sₙ = u₁ + u₂ + … + uₙ is the sum of the first n terms, and S₁, S₂, S₃, … is called the sequence of partial sums.
Recurrence relation: A rule uₙ₊₁ = f(uₙ) that gives each term from the term before it. It is a complete definition only when a starting value, usually u₁, is given.
Sum to infinity: For any series, S∞ is the limit of Sₙ as n → ∞. The series converges if this limit is a finite number. If Sₙ increases without bound or oscillates, the series diverges and the sum to infinity does not exist.
Method of differences: A summation method, also called a telescoping sum, used when the general term can be written as f(r) − f(r + 1). When the rows are written out, the second piece of each row cancels with the first piece of the row below, so only the first and last pieces remain.

Key formulas

Term from sum

$u_n = S_n - S_{n-1} \text{ for } n \geq 2, \quad u_1 = S_1$

Valid only for n ≥ 2, because S₀ is not defined. Find u₁ = S₁ separately and check whether it fits the formula for uₙ.

Arithmetic progression

$u_n = a + (n-1)d, \quad S_n = \frac{n}{2}\left[2a + (n-1)d\right] = \frac{n}{2}(a + l)$

a is the first term, d the common difference and l the last term. Use the first sum form when a and d are known, and the second when the first and last terms are known. An AP with d ≠ 0 never has a sum to infinity.

Geometric progression

$u_n = ar^{n-1}, \quad S_n = \frac{a(1-r^n)}{1-r} = \frac{a(r^n-1)}{r-1}, \; r \neq 1$

a is the first term (a ≠ 0) and r the common ratio. To keep the denominator positive, use the first form when |r| < 1 and the second when r > 1. If r = 1, Sₙ = na.

Sum to infinity of a GP

$S_\infty = \frac{a}{1-r}, \quad |r| < 1$

Exists only when −1 < r < 1, because then rⁿ → 0. State that |r| < 1 before using it. For every other value of r the series diverges.

Sum from the mth term to the nth term

$\sum_{r=m}^{n} u_r = S_n - S_{m-1}$

Subtract Sₘ₋₁, not Sₘ, because uₘ must be kept. There are n − m + 1 terms.

Method of differences

$\sum_{r=1}^{n} \left[f(r) - f(r+1)\right] = f(1) - f(n+1)$

Both pieces must come from the same function f. For a two-step gap, f(r) − f(r + 2), two terms survive at each end: f(1) + f(2) − f(n + 1) − f(n + 2).

Worked example

(i) Express 2 ÷ [(2r − 1)(2r + 1)] in partial fractions. (ii) Hence find Σ (r = 1 to n) 2 ÷ [(2r − 1)(2r + 1)]. (iii) Explain why the series converges and state the sum to infinity. (iv) Find Σ (r = 6 to n) 2 ÷ [(2r − 1)(2r + 1)].

1Write 2 ÷ [(2r − 1)(2r + 1)] = A ÷ (2r − 1) + B ÷ (2r + 1), then multiply through by (2r − 1)(2r + 1) to get 2 = A(2r + 1) + B(2r − 1).
2Substituting r = 1 ÷ 2 gives 2 = 2A, so A = 1. Substituting r = −1 ÷ 2 gives 2 = −2B, so B = −1.
3Hence 2 ÷ [(2r − 1)(2r + 1)] = 1 ÷ (2r − 1) − 1 ÷ (2r + 1).
4Write out the rows of the sum without simplifying them: r = 1 gives 1 − 1 ÷ 3, r = 2 gives 1 ÷ 3 − 1 ÷ 5, r = 3 gives 1 ÷ 5 − 1 ÷ 7, …, r = n − 1 gives 1 ÷ (2n − 3) − 1 ÷ (2n − 1), and r = n gives 1 ÷ (2n − 1) − 1 ÷ (2n + 1).
5Each negative piece cancels with the positive piece in the row below, so only 1 and −1 ÷ (2n + 1) survive. This gives Sₙ = 1 − 1 ÷ (2n + 1) = 2n ÷ (2n + 1).
6As n → ∞, 1 ÷ (2n + 1) → 0, so Sₙ → 1, which is a finite constant. Hence the series converges and the sum to infinity is 1.
7The sum from r = 6 removes the first five terms, so it equals Sₙ − S₅ = [1 − 1 ÷ (2n + 1)] − [1 − 1 ÷ 11] = 1 ÷ 11 − 1 ÷ (2n + 1).

Watch out: For the convergence explanation, use the form 1 − 1 ÷ (2n + 1), not 2n ÷ (2n + 1). In the first form it is obvious which term tends to zero. In the second, both the numerator and the denominator grow, so the limit cannot be read off directly.

Common mistakes

Forgetting to check u₁

Wrong: Finding uₙ = Sₙ − Sₙ₋₁ and stopping there.

Right: That formula only holds for n ≥ 2. Work out u₁ = S₁ separately and check that it agrees with uₙ when n = 1. If it does not, state the first term on its own.

Quoting S∞ with no condition

Wrong: Writing S∞ = a ÷ (1 − r) for any GP.

Right: The sum to infinity exists only when |r| < 1. State this condition, using the actual value of r, before you use the formula. If |r| ≥ 1, S∞ does not exist.

Wrong cancelling with a two-step gap

Wrong: For a general term such as 1 ÷ r − 1 ÷ (r + 2), keeping only one term at the front and one at the back.

Right: Terms cancel with the row two lines below, so two terms survive at each end. Write out the first three rows and the last three rows in a column before cancelling.

Questions students ask

How do you find the nth term from the sum of the first n terms?

For n ≥ 2, use uₙ = Sₙ − Sₙ₋₁. Replace n by (n − 1) in Sₙ, putting brackets around every (n − 1). Then find u₁ = S₁ separately and check that it fits: for Sₙ = 3n² + 2n, uₙ = 6n − 1 and u₁ = S₁ = 5 agrees, so the formula holds for all n ≥ 1.

When does a geometric series have a sum to infinity?

Only when |r| < 1, that is −1 < r < 1. Then rⁿ → 0 and Sₙ → a ÷ (1 − r). For r = 1, r = −1 or |r| > 1 the series diverges and S∞ does not exist. The condition is on |r|, so a negative ratio such as r = −2 ÷ 3 still gives a convergent series.

How do you prove a sequence is an AP or a GP?

For an AP, work out uₙ − uₙ₋₁ for a general n and simplify; for a GP, work out uₙ ÷ uₙ₋₁. If the result is a constant independent of n, the sequence is an AP (or GP), and that constant is the common difference (or ratio). Checking only the first few terms, such as u₂ − u₁ = u₃ − u₂, is not a proof.

Can I use the GC table to find the least value of n in a series question?

Yes, but your written answer must show three things: the expression in n that you entered, two consecutive values of n on either side of the threshold with their values, and a final statement of the least n. Writing only ‘from GC, n = 17’ does not earn the method mark.

In the full PDF

•Figures showing a sequence as discrete points and terms approaching a limit
•GC methods for generating terms: the ANS key and SEQ mode, including the index shift
•The full behaviour vocabulary table: increasing, decreasing, constant, alternating, oscillating, convergent, divergent
•Further worked examples on recurrence limits, AP/GP proofs, sigma notation and two-step-gap method of differences
•The remaining common mistakes and mark-winning exam technique, including a model GC-table answer
•A one-page summary of the five key ideas