How do you do this question with the given power series?
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Step-by-step explanation:
tan⁻¹(x) = ∑ₙ₌₀°° (-1)ⁿ x²ⁿ⁺¹ / (2n+1)
tan⁻¹(1/√3) = ∑ₙ₌₀°° (-1)ⁿ (1/√3)²ⁿ⁺¹ / (2n+1)
tan⁻¹(1/√3) = ∑ₙ₌₀°° (-1)ⁿ (1/√3) (1/√3)²ⁿ / (2n+1)
tan⁻¹(1/√3) = (1/√3) ∑ₙ₌₀°° (-1)ⁿ (1/3)ⁿ / (2n+1)
π/6 = (1/√3) ∑ₙ₌₀°° (-1)ⁿ (1/3)ⁿ / (2n+1)
π = (6/√3) ∑ₙ₌₀°° (-1)ⁿ (1/3)ⁿ / (2n+1)
π = 2√3 ∑ₙ₌₀°° (-1)ⁿ / (3ⁿ (2n+1))