SUM OF INFINITE GEOMETRIC SERIES

Examples 1-6 : Find the sum of the following infinite geometric series.

Example 1 :

3 + 1 + 1/3 +.......... ∞

Solution :

a₁ = 3

r = a₂/a₁ = 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 = 3 and r = 1/3.

S_∞ = 3/(1 - 1/3)

= 3/(2/3)

= 3(3/2)

= 9/2

Example 2 :

-3 - 12 - 48 +.......... ∞

Solution :

a₁ = 3

r = a₂/a₁ = -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 :

a₁ = 1

r = a₂/a₁ = -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 :

a₁ = 1

r = a₂/a₁ = (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_∞ = a₁/(1 - r)

Substitute a₁ = 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 :

a₁ = 1

r = a₂/a₁ = -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_∞ = a₁/(1 - r)

Substitute a₁ = 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,

a₁ = 5(-1/5)⁰

= 5(1)

= 5

When k = 2,

a₂ = 5(-1/5)^2-1

= 5(-1/5)¹

= 5(-1/5)

= -1

When k = 3,

a₃ = 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 + .......... ∞

a₁ = 5

r = a₂/a₁ = -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_∞ = a₁/(1 - r)

Substitute a₁ = 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 :

a₁ = 1 and S_∞ = 1.25

Solution :

S_∞ = 1.25

a₁/(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 :

a₁ = -4 and S_∞ = -16/5

Solution :

S_∞ = -16/5

a₁/(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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