Solution: The recurrence $ a_{n+1} = a_n - rac{a_n^3}{6} $ resembles the Taylor series for $ rctan(u) $, where $ rac{d}{du} rctan(u) = rac{1}{1 + u^2} $. However, the recurrence is not exact. Assume the limit $ L $ exists. Then $ L = L - rac{L^3}{6} \Rightarrow rac{L^3}{6} = 0 \Rightarrow L = 0 $. To confirm convergence, note $ a_1 = \pi/2 pprox 1.57 > 1 $, and $ a_{n+1} = a_n(1 - rac{a_n^2}{6}) $. Since $ a_1 < \sqrt{6} $, $ a_n $ is decreasing and bounded below by 0. By monotone convergence, $ a_n o 0 $.

Sat, 02 May 2026 10:54:33 +0000

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