回覆 3# Hawlee 的帖子
計算第 3 題
S_(n+1) = a_1 + a_2 + ... + a_(n+1) ≦ ra_n = r[S_n - S_(n-1)]
S_n ≧ S_(n-1) + [S_(n+1)/r] ≧ 2√[S_(n-1)S_(n+1)/r]
(S_n)^2 ≧ (4/r)S_(n-1)S_(n+1)
S_(n-1)S_(n+1) ≦ (r/4)(S_n)^2
S_(n+1)/S_n ≦ (r/4)[S_n/S_(n-1)] ≦ [(r/4)^2][S_(n-1)/S_(n-2)] ≦ ... ≦ [(r/4)^(n-1)](S_2/S_1)
若 0 < r < 4,當 n → ∞,[(r/4)^(n-1)](S_2/S_1) = 0,不合
故 r ≧ 4