Infinite Geometric Series

Infinite Geometric Series

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

Formula

S = a1/(1 - r). a1: First term, r: Common ratio (-1 < r < 1)

The infinite series, S, means the infinite sum:
S = a1 + a2 + a3 + ....

For a geometric sequence
whose common ratio [r] is [-1 < r < 1],
the infinite geometric series, S, is
S = a1/(1 - r).

Geometric series

Example 1

Find the value of the given series. 1 + 1/2 + 1/4 + 1/8 + ...

The first term is 1.
So a1 = 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.

a1 = 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

Find the value of the given series. The sum of 6*(1/7)^n as n goes from 1 to infinite.

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

So the given summation is an infinite series.

Sigma notation

an = 6⋅(1/7)n

So a1 = 6⋅(1/7).

So a1 = 6/7.

an = 6⋅(1/7)n

As n increases +1,
an changes ×(1/7).

So the given summation is
an infinite geometric series.

So r = 1/7.

Geometric sequence

a1 = 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

Find the value of the given series. 6/7 + (6/7)(-2/3) + (6/7)(-2/3)^2 + (6/7)(-2/3)^3 ...

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

So a1 = 10.
And r = -2/3.

a1 = 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.