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
Example
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
So 2 is the answer.
Example 2
Example
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
So 6 is the answer.
Example 3
Example
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.
So 28/3 is the answer.