Calculus II: Geometric series

1. Geometric series

A geometric series is a series in which each term is equal to the product of its previous term by a fixed common ratio.

That is a series of the form:

Σ an =	a1 + a2 + ... + an + ...
n → ∞

where:
a₂ = r a₁
a₃ = r a₂ = r² a₁
a₄ = r a₂ = r³ a₁
... = ...
a_n = r a_n-1 = rⁿ a₁

The sum of the first n terms is:
s_n = a₁ + a₂ + ... + a_n
= a₁[1 + r + r² + ... + r^{n - 1}] = a₁ Σ r^{k - 1}
from k = 1 to n

with an index shift the geometric series becomes:
s_n = a₁ Σ r^k
from k = 0 to n - 1

sn = Σ ak =	a1 + a2 + ... + an
k: 1 → n

= a1 Σ rk - 1 =	a1 Σ rk
k: 1 → n	k: 0 → n - 1

To determine whether a series is geometric, we evaluate the ratio a_n+1/a_n. If this ratio is constant, the series is geometric.

2. Examples

2.1. Example 1

The sequence {a_n} = {1, 2, 4, 8, ..., 2^{n - 1}, ... } is geometric. Indeed:

a_n+1/a_n = 2ⁿ/(2^{n - 1}) = 2. This ratio is constant.

The series {s_n} = {1 + 2 + 4 + 8 + ... + 2^{n - 1} + ... } is geometric . Indeed:

s_n+1/s_n = 2ⁿ)/(2^{n - 1}) = 2. This ratio is constant.

The ratio is equal to 2 and the first term a_n = 1.

Therefore:

The sun of the n first terms is:

sn = 1 + 2 + 4 + 8 + ... + 2k - 1 =	1 Σ 2k - 1
k: 1 → n	k: 1 → n

The series is:

1 + 2 + 4 + 8 + ... + 2n - 1 + ... =	Σ2k - 1
	k: 1 → ∞

2.2. Example 2

The series 5 + 15 + 45 + ...
Is a geometric series of ratio 3 and first term 5. So

5 + 15 + 45 + ... =	5 Σ 3k - 1
	k: 1 → ∞

3. Convergence of Geometric series

The geometric series:
a₁ + a₂ + ... + a_n + ...
has the general term:

s_n = a₁ + a₂ + ... + a_n.
= a₁[1 + r + r² + ... + r^{n - 1}] = a₁ Σ r^{k - 1}
k from 1 to n

Let's multiply by r the expression of s_n. We find:
r s_n = a₁[r + r + r³ + ... + rⁿ]
Subtracting this equation from the expression of s_n, we find:
s_n - r s_n =
= a₁[1 + r + r² + ... + r^{n - 1}] - a₁[r + r + r³ + ... + rⁿ] =
a₁[ 1 - rⁿ]

Therefore :

s_n - r s_n = a₁[ 1 - rⁿ]
or
s_n = a₁[ 1 - rⁿ]/(1 - r)

s_n = a₁[1 - rⁿ]/(1 - r)

If r > 1
lim rⁿ = + ∞
n → + ∞

If r < 1
lim rⁿ = 0
n → + ∞

If r > 1
lim s_n = + ∞. The series is divergent.
n → + ∞

If r < 1
lim s_n = a₁/(1 - r) . The series is convergent.
n → + ∞

If r = 1
s_n = a₁[1 + r + r² + ... + r^{n - 1}] = n a₁
so
lim s_n = + ∞. The series is divergent.
n → + ∞

If |r| >= 1
lim s_n = + ∞. The series is divergent.
n → + ∞

If |r| < 1
lim s_n = a₁/(1 - r) . The series is convergent.
n → + ∞