Use the power series for tan^-1(x) to prove the following series for pi as the sum of an infinite series

Question

π = 2√(3)∑(n=0 to ∞)(-1)ⁿ/[(2n+1)(3ⁿ)]

Answer 1

Tan^-1x = x - x³/3 + x⁵/5 - x⁷/7 + ... = ∑_{(n = 0 to ∞)} (-1)ⁿ [(1 / (2n+1))x²ⁿ⁺¹], -1 ≤ x ≤ 1

Note: x²ⁿ⁺¹ = x(x²ⁿ) = x(x²)ⁿ

So we have Tan^-1x = ∑_{(n = 0 to ∞)} (-1)ⁿ[(1 / (2n+1))x(x²)ⁿ], -1 ≤ x ≤ 1

Let x = 1/√3. Then π/6 = Tan^-1(1/√3) = (1/√3)∑_{(n = 0 to ∞)} (-1)ⁿ 1 / [(2n+1)(3)ⁿ]

π = (6/√3)∑_{(n=0 to ∞)} (-1)ⁿ 1 / [(2n+1)(3)ⁿ] = 2√3∑_{(n=0 to ∞)} (-1)ⁿ 1 / [(2n+1)(3)ⁿ]