回覆 14# duncan0804 的帖子
19 P
\( x_{k+1} = \frac{1}{3} x_k^2 + x_k \ge x_k \)
故 \( x_n \ge x_1 = \frac{1}{20} \), \( \forall n \) (因遞增)
因此有 \( x_{n+1} = \frac{1}{3} x_n^2 + x_n \ge \frac{1}{1200} +x_n \)
可推得 \( x_{2026} \ge \frac{2025}{1200} +x_1 > 1 \)
