Formula to find the sum of infinite geometric series :
where -1 < r < 1
In the formula above, a_{1} is the first term of the series and r is the common ratio.
r = second term/first term
or
r = a_{2}/a_{1}
Note :
In an infinite geometric series, if the value of r is not in the interval -1 < r < 1, then the sum does not exist.
Examples 1-6 : Find the sum of the following infinite geometric series.
Example 1 :
3 + 1 + 1/3 +.......... ∞
Solution :
a_{1} = 3
r = a_{2}/a_{1} = 1/3
The value of r (= 1/3) is in the interval -1 < r < 1.
So, the sum for the given infinite geometric series exists.
Formula to find the sum of infinite geometric series :
S_{∞} = a_{1}/(1 - r)
Substitute a_{1} = 3 and r = 1/3.
S_{∞} = 3/(1 - 1/3)
= 3/(2/3)
= 3(3/2)
= 9/2
Example 2 :
-3 - 12 - 48 +.......... ∞
Solution :
a_{1} = 3
r = a_{2}/a_{1} = -12/(-3) = 4
Since the value of r (= 4) is not in the interval -1 < r < 1, the sum for the given infinite geometric series does not exist.
Example 3 :
1 - 3 - 48 +.......... ∞
Solution :
a_{1} = 1
r = a_{2}/a_{1} = -3/1 = -3
Since the value of r (= -3) is not in the interval -1 < r < 1, the sum for the given infinite geometric series does not exist.
Example 4 :
1 + 1/2 + 1/4 +.......... ∞
Solution :
a_{1} = 1
r = a_{2}/a_{1} = (1/2)/1 = 1/2
The value of r (= 1/2) is in the interval -1 < r < 1.
So, the sum for the given infinite geometric series exists.
Formula to find the sum of infinite geometric series :
S_{∞} = a_{1}/(1 - r)
Substitute a_{1} = 1 and r = 1/2.
S_{∞} = 1/(1 - 1/2)
= 1/(1/2)
= 1(2/1)
= 2
Example 5 :
1 - 0.5 + 0.25 + .......... ∞
Solution :
a_{1} = 1
r = a_{2}/a_{1} = -0.5/1 = -0.5
The value of r (= -0.5) is in the interval -1 < r < 1.
So, the sum for the given infinite geometric series exists.
Formula to find the sum of infinite geometric series :
S_{∞} = a_{1}/(1 - r)
Substitute a_{1} = 1 and r = -0.5.
S_{∞} = 1/[1 - (-0.5)]
= 1/(1 + 0.5)
= 1/(1.5)
= 1/(3/2)
= 1(2/3)
= 2/3
Example 6 :
_{k = 1}∑^{k = }^{∞}^{ }5(-1/5)^{k-1}
Solution :
Let a_{k} = 5(-1/5)^{k-1}.
When k = 1,
a_{1} = 5(-1/5)^{0}
= 5(1)
= 5
When k = 2,
a_{2} = 5(-1/5)^{2-1}
= 5(-1/5)^{1}
= 5(-1/5)
= -1
When k = 3,
a_{3} = 5(-1/5)^{3-1}
= 5(-1/5)^{2}
= 5(1/25)
= 1/5
_{k = 1}∑^{k = }^{∞}^{ }5(-1/5)^{k-1 }= 5 - 1 + 1/5 + .......... ∞
a_{1} = 5
r = a_{2}/a_{1} = -1/5
The value of r (= -1/5) is in the interval -1 < r < 1.
So, the sum for the given infinite geometric series exists.
Formula to find the sum of infinite geometric series :
S_{∞} = a_{1}/(1 - r)
Substitute a_{1} = 5 and r = -1/5.
S_{∞} = 5/[1 - (-1/5)]
= 5/(1 + 1/5)
= 5/(6/5)
= 5(5/6)
= 25/6
Examples 7-8 : Determine the common ratio of the infinite geometric series.
Example 7 :
a_{1} = 1 and S_{∞} = 1.25
Solution :
S_{∞} = 1.25
a_{1}/(1 - r) = 1.25
1/(1 - r) = 1.25
Take reciprocal on both sides.
(1 - r)/1 = 1/1.25
1 - r = 0.8
Subtract 1 from both sides.
-r = -0.2
Multiply both sides by -1.
r = 0.2
Example 8 :
a_{1} = -4 and S_{∞} = -16/5
Solution :
S_{∞} = -16/5
a_{1}/(1 - r) = -16/5
-4/(1 - r) = -16/5
Take reciprocal on both sides.
(1 - r)/(-4) = -5/16
Multiply both sides by -4.
1 - r = 5/4
Subtract 1 from both sides.
-r = 1/4
Multiply both sides by -1.
r = -1/4
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