Sum the infinite series:

1/5 × (1/5 − 1/7) + (1/5)² × ((1/5)² − (1/7)²) + (1/5)³ × ((1/5)³ − (1/7)³) + …

Question

Sum the infinite series:

1/5 × (1/5 − 1/7) + (1/5)² × ((1/5)² − (1/7)²) + (1/5)³ × ((1/5)³ − (1/7)³) + …

UPSC Quizroom · Accepted Answer

Factor each term:

n-th term = (1/5)ⁿ × ((1/5)ⁿ − (1/7)ⁿ) = (1/5)²ⁿ − (1/5)ⁿ × (1/7)ⁿ = (1/25)ⁿ − (1/35)ⁿ

Rewrite series as difference of two geometric series:

First series: Σ (1/25)ⁿ, n = 1 to ∞ → sum = (1/25) / (1 − 1/25) = 1/24

Second series: Σ (1/35)ⁿ, n = 1 to ∞ → sum = (1/35) / (1 − 1/35) = 1/34

Total sum = 1/24 − 1/34 = (34 − 24) / (24 × 34) = 10 / 816 = 5 / 408

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