Sum of Infinite Geometric Series

An infinite geometric series is a specific type of infinite series where each term after the first is found by multiplying the previous term by a constant called the common ratio. An infinite geometric series is an infinite sum of the form:

S = a + ar + ar² + ar³ + ar⁴ + . . .

Where:

• S is the sum of the series.
• a is the first term.
• r is the common ratio (the factor by which we multiply to get from one term to the next).

For a geometric series given above, we can express the sum as,

a + ar + ar² + ar³ + . . . = a/(1 - r)

Condition for Convergence for Infinite Geometric Series

• The series should be in geometric progression.
• The absolute value of the common ratio should be less than 1.

Derivation of the Formula

Let's consider,

• a = first term of the geometric series
• r = common ratio, where -1 < r < 1

Let us consider the sum of the geometric progression be S.

Then we can write,

S = a + ar + ar²+ ar³ + . . . ---(i)

Multiplying both sides of the equation by r, we get,

Sr = ar + ar² + ar³ + ar⁴ + . . . ---(ii)

Subtracting Eq. (ii) from Eq. (i), we get

S - Sr = (a + ar + ar²+ ar³) + . . . - (ar + ar² + ar³ + ar⁴ + . . . )

S(1 - r) = a

S = a/(1 - r)

Hence, the sum of infinite series of a geometric progression is a/(1 - r)

Note: If the absolute value of the common ratio 'r' is greater than 1, then the sum will not converge.

Thus, the absolute value of the sum will tend to infinity. Thus, if r > 1,

| S | = | a + ar + ar² + ar³ + ... | = ∞

Sample Questions on Sum of Infinite Series

Question 1. Find the sum of the infinite series with first term 4 and common ratio 1/2.

Solution:

Given, the first term a = 4

The common ratio r = 1/2

Thus, we can write the series as,

S = 4 + 4 × (1/2) + 4 × (1/2)² + ...

So, the sum will stand as

S = 4/(1 - (1 / 2)) = 4/(1/2) = 4 × 2 = 8
⇒ S = 8

So, the sum of the series is equal to 8.

Question 2. Find the sum of the infinite series 1 + (1/2) + (1/2)² + (1/2)³ + ... .

Solution:

Given, the first term of the series a =  1.

The common ratio is  r = 1/2.

Since the absolute value of the common ratio is less than 1, we can apply the general formula.

So, the sum is,

S = 1/(1 - (1/2)) = 2

So, the sum of the given infinite series is 2.

Question 3. Evaluate the sum 2 + 4 + 8 + 16 + ... .

Solution:

We can write the sum of the given series as,

S = 2 + 2² + 2³ + 2⁴ + ...

We can observe that it is a geometric progression with infinite terms and first term equal to 2 and common ratio equals 2.

Thus, r = 2.

Since, the value of r > 1, the sum will not converge and tend to infinity. Thus,

S = + ∞

Question 4. Find the sum of the series 2 - 1/5 + 1 - 1/25 + 1/2 - 1/125 + ... .

Solution:

We can write the sum of the series as the difference of two infinite series as:

S = (2 + 1 + 1/2 + 1/2² + . . . ) - (1/5 + 1/25 + 1/125 + ... )
⇒ S = (2 + 1 + 1/2 + 1/2² + . . . ) - (1/5 + 1/5² + 1/5³ + ...)

Let S = S₁ - S₂

Where,

• S₁ = 2 + 1 + 1/2 + 1/2² + . . .
• S₂ = 1/5 + 1/5² + 1/5³ + . . .

Here, we can see both S₁ and S₂ are infinite summation of geometric series, where,

• a₁ = 2, r₁ = 1/2
• a₂ = 1/5, r₂ = 1/5

Thus, we can write,

• S₁ = 2/(1 - (1/2)) = 2/(1/2) = 4
• S₂ = (1/5)/(1 - (1/5)) = (1/5) / (4/5) = 1/4

So, the summation S stands as,

S = S1 - S2 = 4 - 1/4 = (16 - 1)/4 = 15/4 = 3.75
⇒ S = 3.75

Thus, the sum of the given series is 3.75 .

Read More,

• Sequence and Series
• Geometric Series
• Arithmetic Series

• If ∣r∣ ≥ 1, the series diverges, meaning the sum does not approach a finite limit.