Q.No. 51 An infinite series S is given as:

S = 1 + 2/3 + 3/9 + 4/27 + 5/81 + … (to infinity)

The value of S is _____________ (round off to 2 decimal places).

Infinite Series Sum 1 + 2/3 + 3/9 + 4/27 + 5/81 to Infinity | Value 2.25 Rounded

The infinite series S = 1 + 2/3 + 3/9 + 4/27 + 5/81 + ⋯ is an arithmetico-geometric series with general term
a_n = n / 3ⁿ⁻¹. Its sum converges to 2.25 (9/4).

Series Identification

The numerators form an arithmetic sequence: 1, 2, 3, 4, 5, …

The denominators form a geometric sequence: 1 = 3⁰, 3 = 3¹, 9 = 3², 27 = 3³, 81 = 3⁴, …

Thus, the nth term is u_n = n(1/3)ⁿ⁻¹, confirming convergence since |r| = 1/3 < 1.

Sum Derivation

Consider the infinite sum S = ∑ n=1→∞ n rⁿ⁻¹ where r = 1/3.

Start from geometric series:
∑ n=0→∞ rⁿ = 1 / (1 − r)

Differentiating both sides with respect to r gives:
∑ n=1→∞ n rⁿ⁻¹ = 1 / (1 − r)²

Substitute r = 1/3:

S = 1 / (1 − 1/3)² = 1 / (2/3)² = 1 / (4/9) = 9/4 = 2.25

Verification by Partial Sums

• S₁ = 1
• S₂ = 1 + 0.6667 = 1.6667
• S₃ = 1.6667 + 0.3333 = 2.0000
• S₄ = 2.0000 + 0.1481 = 2.1481
• S₅ = 2.1481 + 0.0617 = 2.2099

The values approach 2.25, confirming convergence.

Introduction to Infinite Series Sum

The infinite series sum S = 1 + 2/3 + 3/9 + 4/27 + 5/81 + ⋯ appears in competitive exams like IIT JAM and GATE. This arithmetico-geometric progression converges to 9/4 = 2.25, which is the correct total.

Step-by-Step Solution

General Term: u_n = n(1/3)ⁿ⁻¹

Base Formula:
∑ rⁿ = 1 / (1 − r)
Differentiate → ∑ n rⁿ⁻¹ = 1 / (1 − r)²

Plug r = 1/3: S = (2/3)⁻² = 2.25

Common mistakes include treating the series as purely geometric (sum = 1.5), or forgetting the derivative step.

Exam Relevance

Ideal for IIT JAM Mathematics—verify convergence with partial sums and apply differentiation for arithmetico-geometric series.