Define S_n = Σ (from k = 1 to n) of 3^(k−1)·sin³(π/3^k). It is known…

Question

Define S_n = Σ (from k = 1 to n) of 3^(k−1)·sin³(π/3^k). It is known that S_n converges as n → ∞. Let L be that limit. Compute ⌊10⁴·L⌋ (the floor of ten thousand times L).

Accepted Answer

Use the triple-angle identity sin 3A = 3 sin A − 4 sin³A, which gives sin³A = (3 sin A − sin 3A)/4.

With A = π/3^k: 3^{k−1} sin³(π/3^k) = (3^{k−1}/4)[3 sin(π/3^k) − sin(π/3^{k−1})] = (1/4)[3^k sin(π/3^k) − 3^{k−1} sin(π/3^{k−1})].

This telescopes: S_n = (1/4)[3^n sin(π/3^n) − 3^0 sin(π/3^0)] = (1/4)[3^n sin(π/3^n) − sin π] = (1/4)·3^n sin(π/3^n).

As n → ∞, 3^n sin(π/3^n) → 3^n·(π/3^n) = π (since sin θ ≈ θ for small θ). Hence L = π/4 ≈ 0.7853981634.

Then 10⁴·L ≈ 7853.98, so ⌊10⁴·L⌋ = 7853.

Final answer: 7853

Define S_n = Σ (from k = 1 to n) of 3^(k−1)·sin³(π/3^k). It is known…

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