DETERMINE WHETHER THE GEOMETRIC SERIES IS CONVERGENT OR DIVERGENT
Consider the series
4 + 2 + 1 + 0.5 + 0.25 + .............
s5 = 7.75 (Sum of first 5 terms)
s7= 7.9375 (Sum of first 7 terms)
s9 = 7.9844......... (Sum of first 9 terms)
Convergent series:
As the number of terms increases, the sequence of partial sums approaches a fixed value of 8. Therefore, the sum of this series is 8. This series is said to be a convergent series.
Divergent series:
As the number of terms increases, the sum of the series continues to grow. The sequence of partial sums does not approach a fixed value. Therefore, the sum this series cannot be calculated. This series is said to be a divergent series.
Consider the example,
4 + 8 + 16 + 32 +.............
s2 = 4 (Sum of first 2 terms)
s4 = 60 (Sum of first 4 terms)
s6 = 252 (Sum of first 6 terms)
The sum of an infinite geometric series, where -1 < r < 1, can be determined using the formula
S∞ = t1 / (1 - r)
Where t1 is the first term of the series
r is the common ratio
S∞ represents the sum of an infinite number of terms
Decide whether each infinite geometric series is convergent or divergent. State the sum of the series, if it exists.
Problem 1 :
1 - 1/3 + 1/9 - …
Solution :
a1 = 1
r = a2 / a1
r = (-1/3) / 1
r = -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∞ = a1 / (1 - r)
Substitute a1 = 1 and r = -1/3.
Convergent, the sum exists.
S∞ = a1 / (1 - r)
= 1 / (1 + 1/3)
= 1 / (4/3)
S∞ = 3/4
Problem 2 :
2 - 4 + 8 - …
Solution :
a1 = 2
r = a2 / a1
r = -4 / 2
r = -2
Since the value of r = -2 is not in the interval -1 < r < 1, the sum for the given infinite geometric series does not exists.
State whether each infinite geometric series is convergent or divergent.
Problem 3 :
t1 = -3, r = 4
Solution :
t1 = -3
r = 4
The value of r = 4 is not in the interval -1 < r < 1. So, the given series is divergent.
Problem 4 :
t1 = 4, r = -1/4
Solution :
t1 = 4
r = -1/4
The value of r = -1/4 is in the interval -1 < r < 1. So, the given series is convergent.
Problem 5 :
125 + 25 + 5 + …
Solution :
a1 = 125
r = a2 / a1
r = 25/125
r = 1/5
The value of r = 1/5 is in the interval -1 < r < 1. So, the given series is convergent.
Problem 6 :
(-2) + (-4) + (-8) + …
Solution :
a1 = -2
r = a2 / a1
r = -4/-2
r = 2
The value of r = 2 is not in the interval -1 < r < 1. So, the given series is divergent.
Problem 7 :
243/3125 - 81/625 + 27/25 - 9/5 + ….
Solution :
a1 = 243/3125
r = a2 / a1
r = (-81/625) / (243/3125)
r = -5/3
The value of r = -5/3 is not in the interval -1 < r < 1. So, the given series is divergent.
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