# Infinite Geometric Series

How to find the value of an infinite geometric series: formula, 3 examples, and their solutions.

## Formula

### Formula

An infinite series means
a series when n goes to infinity.

S = a1 + a2 + a3 + ... + an + ...

For an infinite geometric series,
if the common ratio r is -1 < r < 1
(|r| < 1),
then the sum S goes to this value.

S = a/[1 - r]

S: Infinite geometric series
a: First term
r: Common ratio

If |r| < 1 (-1 < r < 1),
as n increases,
an goes to 0.
So the sum Sn = [a(1 - rn)]/[1 - r]
goes to
S = [a(1 - 0)]/[1 - r]
= a/[1 - r].

## Example 1

### Solution

Write the terms of the given series
as an = arn - 1 form.

Geometric Sequence

Write the first term 1.

This is a.

The second term is +1/2.

So write
plus,
a, 1
times,
(+1/2)/1, 1/2
to the first:
+1⋅(1/2)1.

The third term is +1/4.

a = 1

So write +1⋅(1/2)2.

The fourth term is +1/8.

a = 1

So write +1⋅(1/2)3.

Write + ... .

So
1 + 1/2 + 1/4 + 1/8 + ...
= 1 + 1⋅(1/2)1 + 1⋅(1/2)2 + 1⋅(1/2)3 + ... .

a = 1
r = 1/2
n goes to ∞.

So the sum is
1/[1 - 1/2].

Multiply 2
to both of the numerator and the denominator.

1⋅2 = 2

[1 - 1/2]⋅2 = 2 - 1

Common Monomial Factor

2 - 1 = 1

2/1 = 2

## Example 2

### Solution

Write the terms of the given series
as an = arn - 1 form.

Write the first term 10.

This is a.

The second term is -20/3.

So write
plus,
a, 10
times,
(-20/3)/10, -2/3
to the first:
+10⋅(-2/3)1.

The third term is +40/9.

a = 10

So write +10⋅(-2/3)2.

The fourth term is -80/27.

a = 10

So write +10⋅(-2/3)3.

Write + ... .

So
10 - 20/3 + 40/9 - 80/27 + ...
= 10 + 10⋅(-2/3)1 + 10⋅(-2/3)2 + 10⋅(-2/3)3 + ... .

a = 10
r = -2/3
n goes to ∞.

So the sum is
10/[1 - (-2/3)].

-(-2/3) = +2/3

Multiply 3
to both of the numerator and the denominator.

10⋅3 = 30

(1 + 2/3)⋅3 = 3 + 2

3 + 2 = 5

30/5 = 6

## Example 3

### Solution

See the sigma notation.
n goes from 0.

So the first term, a1, is
when n = 0.

So a1 = 8⋅(1/7)0.

(1/7)0 = 1

Zero Exponent

8⋅1 = 8

So a1 = 8.

See the term in the sigma:
8⋅(1/7)n.

As n increases,
(1/7) is multiplied:
8⋅(1/7)0, 8⋅(1/7)1, 8⋅(1/7)2, ... .

So the given summation
is an infinite geometric series.

So r = 1/7.

a1 = a = 8
r = 1/7
n goes to ∞.

So the sum is
8/[1 - 1/7].

Multiply 7
to both of the numerator and the denominator.

8⋅7 = 56

(1 - 7)⋅7 = 7 - 1

7 - 1 = 6

Reduce the numerator 56 to, 56/2, 28
and reduce 6 to, 6/2, 3.