Sequences and Series

A sequence is an arrangement of numbers in a definite order according to some rule. If a₁, a₂, a₃, ..., a_n, ... are the terms of a sequence, then a_n is called the nth term (or general term) of the sequence. A sequence with a finite number of terms is a finite sequence; otherwise it is an infinite sequence.

When the terms of a sequence are added, the expression a₁ + a₂ + a₃ + ... + a_n + ... is called a series. A series is finite or infinite according as the corresponding sequence is finite or infinite. The sum of the first n terms is denoted S_n, and a_n = S_n − S_n−1 for n ≥ 2 (with a₁ = S₁). This relation lets us recover the general term whenever only S_n is known.

Worked Example. Write the first five terms of the sequence whose nth term is a_n = (2n − 3)/6.

Substituting n = 1, 2, 3, 4, 5:

• a₁ = (2(1) − 3)/6 = −1/6
• a₂ = (2(2) − 3)/6 = 1/6
• a₃ = (2(3) − 3)/6 = 3/6 = 1/2
• a₄ = (2(4) − 3)/6 = 5/6
• a₅ = (2(5) − 3)/6 = 7/6

So the sequence is −1/6, 1/6, 1/2, 5/6, 7/6, ... and the associated series is −1/6 + 1/6 + 1/2 + 5/6 + 7/6 + ... .

A sequence a₁, a₂, a₃, ... is an arithmetic progression (AP) if there is a constant d, the common difference, such that a_n+1 = a_n + d for every n. With first term a, the AP is a, a + d, a + 2d, a + 3d, ... .

The nth term is a_n = a + (n − 1)d. If the AP ends at a last term l after n terms, then l = a + (n − 1)d.

The sum of the first n terms is S_n = (n/2)[2a + (n − 1)d], which can equivalently be written as S_n = (n/2)(a + l).

Worked Example 1. Find the sum of the first 20 terms of the AP 2, 7, 12, 17, ... .

Here a = 2, d = 7 − 2 = 5. Using S_n = (n/2)[2a + (n − 1)d] with n = 20:

S₂₀ = (20/2)[2(2) + 19(5)] = 10[4 + 95] = 10(99) = 990.

Worked Example 2. Which term of the AP 3, 8, 13, 18, ... is 78?

Here a = 3, d = 5. Setting a_n = a + (n − 1)d = 78:

3 + (n − 1)(5) = 78 ⟹ (n − 1)(5) = 75 ⟹ n − 1 = 15 ⟹ n = 16. So 78 is the 16th term.

If a, A, b are in AP, A is the arithmetic mean (AM) of a and b, and A = (a + b)/2 (the middle term of an AP equals the average of its neighbours).

More generally, if A₁, A₂, ..., A_n are inserted between a and b so that a, A₁, A₂, ..., A_n, b forms an AP of (n + 2) terms, they are the n arithmetic means between a and b. Treating b as the (n + 2)th term, b = a + (n + 1)d, so d = (b − a)/(n + 1), and the kth mean is A_k = a + kd.

Worked Example. Insert 3 arithmetic means between 3 and 19.

Here a = 3, b = 19, n = 3, so d = (19 − 3)/(3 + 1) = 16/4 = 4.

The means are A₁ = 3 + 4 = 7, A₂ = 7 + 4 = 11, A₃ = 11 + 4 = 15. The resulting AP is 3, 7, 11, 15, 19 — a common difference of 4 throughout.

A useful fact: the sum of the n inserted arithmetic means equals n times the single AM of a and b, i.e., A₁ + A₂ + ... + A_n = n(a + b)/2.

A sequence of non-zero numbers a₁, a₂, a₃, ... is a geometric progression (GP) if there is a constant r ≠ 0, the common ratio, such that a_n+1/a_n = r for every n. With first term a, the GP is a, ar, ar², ar³, ... .

The nth term is a_n = arⁿ⁻¹.

The sum of the first n terms is:

• S_n = a(rⁿ − 1)/(r − 1), equivalently a(1 − rⁿ)/(1 − r), when r ≠ 1;
• S_n = na, when r = 1.

When |r| < 1, rⁿ → 0 as n grows, so the sum to infinity exists: S_∞ = a/(1 − r).

Worked Example 1. Find the sum of the first 8 terms of the GP 3, 6, 12, 24, ... .

Here a = 3, r = 6/3 = 2. Since r > 1, S₈ = a(r⁸ − 1)/(r − 1) = 3(2⁸ − 1)/(2 − 1) = 3(256 − 1) = 3(255) = 765.

Worked Example 2. Find the sum to infinity of the GP 3, 1, 1/3, 1/9, ... .

Here a = 3, r = 1/3, and |r| < 1, so S_∞ = a/(1 − r) = 3/(1 − 1/3) = 3/(2/3) = 9/2.

If a, G, b (all positive) are in GP, G is the geometric mean (GM) of a and b, and G² = ab, so G = √(ab).

More generally, if G₁, G₂, ..., G_n are inserted between a and b (a, b > 0) so that a, G₁, G₂, ..., G_n, b is a GP, they are the n geometric means between a and b. Treating b as the (n + 2)th term, b = arⁿ⁺¹, so r = (b/a)^1/(n+1), and the kth mean is G_k = ar^k.

Worked Example. Insert 2 geometric means between 1 and 8.

Here a = 1, b = 8, n = 2, so r = (8/1)^1/3 = 2. The means are G₁ = 1(2) = 2, G₂ = 1(2²) = 4. The GP is 1, 2, 4, 8, with common ratio 2 throughout.

Relationship between AM and GM. For positive reals a and b, with A = (a + b)/2 and G = √(ab):

A − G = (a + b)/2 − √(ab) = (√a − √b)²/2 ≥ 0, so A ≥ G always, with equality if and only if a = b.

Also, since a + b = 2A and ab = G², the numbers a and b are the two roots of x² − 2Ax + G² = 0.

Worked Example. The AM and GM of two positive numbers are 5 and 4 respectively. Find the numbers.

The numbers are roots of x² − 2(5)x + 4² = 0, i.e., x² − 10x + 16 = 0. So x = [10 ± √(100 − 64)]/2 = [10 ± 6]/2, giving x = 8 or x = 2. Check: AM of 8 and 2 = 5 ✓; GM = √16 = 4 ✓.

Three standard series recur throughout this chapter — the sums of the first n natural numbers, their squares, and their cubes:

• Sum of first n natural numbers: 1 + 2 + 3 + ... + n = Σn = n(n + 1)/2.
• Sum of squares of first n natural numbers: 1² + 2² + ... + n² = Σn² = n(n + 1)(2n + 1)/6.
• Sum of cubes of first n natural numbers: 1³ + 2³ + ... + n³ = Σn³ = [n(n + 1)/2]² = (Σn)².

These are used to sum more complicated series whose nth term is a polynomial in n, by splitting the nth term into simpler pieces (Σn³, Σn², Σn, Σ constant) and summing each piece separately.

Worked Example. Find the sum to n terms of the series 1×2 + 2×3 + 3×4 + ... + n(n + 1).

The nth term is a_n = n(n + 1) = n² + n. So:

S_n = Σn² + Σn = n(n + 1)(2n + 1)/6 + n(n + 1)/2.

Taking n(n + 1)/6 common: S_n = [n(n + 1)/6][(2n + 1) + 3] = [n(n + 1)/6](2n + 4) = n(n + 1)(n + 2)/3.

Check for n = 2: actual sum = 1(2) + 2(3) = 2 + 6 = 8; formula gives 2(3)(4)/3 = 24/3 = 8. ✓