# Infinite Geometric Series

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

## Formula

An infinite series means

a series when n goes to infinity.

S = a_{1} + a_{2} + a_{3} + ... + a_{n} + ...

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,

a_{n} goes to 0.

So the sum S_{n} = [a(1 - r^{n})]/[1 - r]

goes to

S = [a(1 - 0)]/[1 - r]

= a/[1 - r].

## Example1 + 1/2 + 1/4 + 1/8 + ...

Write the terms of the given series

as a_{n} = ar^{n - 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

So 2 is the answer.

## Example10 - 20/3 + 40/9 - 80/27 + ...

Write the terms of the given series

as a_{n} = ar^{n - 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

So 6 is the answer.

## Example∑_{n = 0}^{∞} 8⋅(1/7)^{n}

See the sigma notation.

n goes from 0.

So the first term, a_{1}, is

when n = 0.

So a_{1} = 8⋅(1/7)^{0}.

(1/7)^{0} = 1

Zero Exponent

8⋅1 = 8

So a_{1} = 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.

a_{1} = 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.

So 28/3 is the answer.