# 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)

s= 60 (Sum of first 4 terms)

s= 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

Srepresents 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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