Вычислить сумму ряда с точностью α. Σ_{n=1}^∞ ((-1)^{n+1}) / ((2n)^3), α = 0,001.

The series is an alternating series. We need to find the number of terms such that the absolute value of the last term is less than α.

The general term is a_n = ((-1)^{n+1}) / ((2n)^3). We need |a_n| < 0.001.

|a_n| = 1 / (8n^3) < 0.001 => 8n^3 > 1000 => n^3 > 125 => n > 5. So we need to sum up to n=6.

S_6 = 1/8 * (1 - 1/2^3 + 1/3^3 - 1/4^3 + 1/5^3 - 1/6^3) = 1/8 * (1 - 1/8 + 1/27 - 1/64 + 1/125 - 1/216) ≈ 0.125 * (1 - 0.125 + 0.037037 - 0.015625 + 0.008 - 0.004629) ≈ 0.125 * 0.904783 ≈ 0.113098.