# Infinite Geometric Series

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

## Formula

The infinite series, *S*, means the infinite sum:*S* = *a*_{1} + *a*_{2} + *a*_{3} + ....

For a geometric sequence

whose common ratio [*r*] is [-1 < *r* < 1],

the infinite geometric series, *S*, is*S* = *a*_{1}/(1 - *r*).

Geometric series

## Example 1

The first term is 1.

So *a*_{1} = 1.

See the terms of the given series.

(1/2) / 1 = 1/2

(1/4) / (1/2) = 1/2

(1/8) / (1/4) = 1/2

...

So *r* = 1/2.

*a*_{1} = 1*r* = 1/2

So *S* = 1/(1 - 1/2).

1 - 1/2 = 1/2

Multiply 2

to both of the numerator and the denominator.

Then *S* = 2.

## Example 2

See the given summation.*n* goes from 1 to infinity.

So the given summation is an infinite series.

Sigma notation

*a*_{n} = 6⋅(1/7)^{n}

So *a*_{1} = 6⋅(1/7).

So *a*_{1} = 6/7.

*a*_{n} = 6⋅(1/7)^{n}

As *n* increases +1,*a*_{n} changes ×(1/7).

So the given summation is

an infinite geometric series.

So *r* = 1/7.

Geometric sequence

*a*_{1} = 6/7*r* = 1/7

So *S* = (6/7) / (1 - 1/7).

1 - 1/7 = 6/7

The numerator and the denominator are the same:

6/7.

So *S* = 1.

## Example 3

10 = 10

-20/3 = 10⋅(-2/3)

+40/9 = 10⋅(-2/3)^{2}

-80/27 = 10⋅(-2/3)^{3}

...

So *a*_{1} = 10.

And *r* = -2/3.

*a*_{1} = 10*r* = -2/3

So *S* = 10 / [1 - (-2/3)].

1 - (-2/3) = 3/3 + 2/3

10 = 10/1

3/3 + 2/3 = 5/3

(10/1) / (5/3) = (10⋅3)/(1⋅5)

= 30/5

Complex fraction

30/5 = 6

So *S* = 6.