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H2 Mathematics · JC 1 and 2

Sequences and Series

A sequence is an ordered list of numbers, and a series is the sum of its terms. In H2 Mathematics you work with arithmetic and geometric progressions, sums to infinity, sigma notation, the method of differences and sequences defined by recurrence relations. Questions are often set in context, such as savings plans or loan repayments, and test whether you can set up the right sum as well as compute it. These ideas also prepare you for Maclaurin series later in the course.

What the syllabus expects

  • Describe sequences given by a formula for the nth term or by a recurrence relation, and state whether they converge
  • Use the formulae for the nth term and the sum of the first n terms of arithmetic and geometric progressions
  • Use the condition |r| < 1 for a geometric series to converge, and find its sum to infinity
  • Use sigma notation and its properties, including the standard results for the sums of r and r²
  • Find the sum of a series using the method of differences
  • Solve problems in context, including finding the least n for which a sum exceeds a given value

Arithmetic progressions

An arithmetic progression (AP) has a common difference d between consecutive terms. With first term a, the nth term is u_n = a + (n - 1)d, and the sum of the first n terms is S_n = n/2 [2a + (n - 1)d], or n/2 (a + l) where l is the last term. For example, in 5, 8, 11, ..., the 20th term is 5 + 19 × 3 = 62 and the sum of the first 20 terms is 10(5 + 62) = 670. A useful link for any sequence is u_n = S_n - S_(n-1). If S_n = 2n² + 3n, then u_n = 2n² + 3n - [2(n - 1)² + 3(n - 1)] = 4n + 1, so the sequence is an AP with first term 5 and common difference 4.

Geometric progressions and sums to infinity

A geometric progression (GP) has a common ratio r. The nth term is u_n = ar^(n-1) and, for r ≠ 1, S_n = a(1 - r^n)/(1 - r). If |r| < 1, then r^n gets closer to 0 as n increases, so the series converges and the sum to infinity is S∞ = a/(1 - r). If |r| ≥ 1, the series has no sum to infinity. For example, 24 + 12 + 6 + ... has a = 24 and r = 1/2, so S∞ = 24/(1 - 1/2) = 48. Questions about the least n often lead to an inequality involving r^n. Solve it by taking logarithms, remembering that ln r is negative when 0 < r < 1, which reverses the inequality, or use a table of values on the graphing calculator.

Sigma notation and the method of differences

Sigma notation writes a sum compactly: Σ r² for r = 1 to n means 1² + 2² + ... + n². Constants factor out and sums split, so Σ(3r² + 2) for r = 1 to n equals 3Σr² + 2n. Standard results include Σr = n(n + 1)/2 and Σr² = n(n + 1)(2n + 1)/6, both for r = 1 to n. Watch the starting value: a sum from r = 5 to n equals the sum from 1 to n minus the sum from 1 to 4. The method of differences works when each term can be written as f(r) - f(r + 1), often after partial fractions. Writing out the first few and last few rows shows that most terms cancel. For example, 1/(r(r + 1)) = 1/r - 1/(r + 1), so the sum from r = 1 to n is 1 - 1/(n + 1). As n becomes large this tends to 1, so the infinite series converges to 1.

Recurrence relations and convergence

A recurrence relation defines each term from the one before, such as u_(n+1) = ½u_n + 3 with u_1 = 2. Generating terms gives 2, 4, 5, 5.5, 5.75 and so on. If the sequence converges to a limit L, then for large n both u_n and u_(n+1) are close to L, so L = ½L + 3, which gives L = 6. A sequence diverges if its terms do not approach a single finite value, for example if they grow without bound or keep oscillating. The terms of an AP with d ≠ 0 diverge, while the terms of a GP with |r| < 1 converge to 0. Be precise with words: a sequence converges when its terms approach a limit, and a series converges when its partial sums S_n approach a limit.

Worked example

At the start of every month, Mei deposits $500 into a savings account. At the end of every month, interest of 0.5% of the amount in the account is added. (a) Show that the amount in the account at the end of the nth month is 100500(1.005^n - 1) dollars. (b) Find the least number of months for the amount to exceed $20 000.

  1. End of month 1: 500(1.005). End of month 2: (500(1.005) + 500) × 1.005 =500(1.005² + 1.005).
  2. Following the pattern, the end of month n gives 500(1.005 + 1.005² + ... + 1.005^n).
  3. The bracket is a GP with first term 1.005, ratio 1.005 and n terms, so its sum is 1.005(1.005^n - 1)/(1.005 - 1) = 201(1.005^n - 1).
  4. So the amount is 500 × 201(1.005^n - 1) = 100500(1.005^n - 1) dollars, as required.
  5. For the amount to exceed $20 000, 100500(1.005^n - 1) > 20000, so 1.005^n > 1 + 20000/100500 = 1.19900 to 5 decimal places.
  6. Take natural logarithms: n ln 1.005 > ln 1.19900. Since ln 1.005 > 0, n > 36.4 to 3 significant figures.
  7. Check: n = 36 gives about $19 766 and n = 37 gives about $20 368, so the least n is 37.

Answer: 37 months

Common mistakes

  • Using S∞ = a/(1 - r) without checking that |r| < 1. State the condition before using the formula.
  • Miscounting the terms in a sigma sum. The sum from r = 3 to r = 10 has 8 terms, not 7 or 10.
  • Dividing by ln r without reversing the inequality when 0 < r < 1. Since ln r is negative there, the inequality sign flips.
  • Treating Σ(3r + 2) for r = 1 to n as 3Σr + 2. The constant is added once for every term, so it contributes 2n.
  • Cancelling the wrong terms in the method of differences. Write out at least the first three and last three rows in columns so that the surviving terms are clear.
  • In money problems, mixing up deposits at the start and at the end of a month, which shifts every power by one. Write out the first two months in full before generalising.

Exam tips

  • For least n questions, show a table from the graphing calculator with the values either side of the boundary. It gives a clear, checkable conclusion.
  • In method of differences questions, line up cancelling terms vertically. The pattern of what survives is easier to see, especially when terms cancel two rows apart.
  • In context questions, write the first two or three stages in full before generalising to n. It fixes the powers and the number of terms.

Practice questions

1. The third term of a geometric progression is 18 and the sixth term is 486. Find the first term and the common ratio.

Show answer

a = 2 and r = 3. From ar² = 18 and ar⁵ = 486, dividing gives r³ = 27, so r = 3. Then a = 18/9 = 2.

2. Find the value of Σ(3r² - 2r) for r = 1 to 20.

Show answer

8190. Σr² for r = 1 to 20 is 20 × 21 × 41/6 = 2870, and Σr is 20 × 21/2 = 210. So the sum is 3(2870) - 2(210) = 8610 - 420 = 8190.

3. Show that 1/r - 1/(r + 2) = 2/(r(r + 2)). Hence find Σ 2/(r(r + 2)) for r = 1 to n, and deduce the sum to infinity.

Show answer

Over a common denominator, 1/r - 1/(r + 2) = (r + 2 - r)/(r(r + 2)) = 2/(r(r + 2)). Writing out the rows (1 - 1/3) + (1/2 - 1/4) + (1/3 - 1/5) + ... + (1/(n - 1) - 1/(n + 1)) + (1/n - 1/(n + 2)), the terms cancel in pairs two rows apart, leaving 3/2 - 1/(n + 1) - 1/(n + 2). As n becomes large the last two terms tend to 0, so the sum to infinity is 3/2.

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