olymmath

DE-MO 2002 Problem 4

    Given a positive real number a1a_1 , we recursively define an+1=1+a1a2⋯⋅an.a_{n+1} = 1+a_1 a_2 \cdots \cdot a_n. Furthermore, let bn=1a1+1a2+⋯+1an.b_n = \frac{1}{a_1 } + \frac{1}{a_2 } +\cdots + \frac{1}{a_n }. Prove that bn<2a1b_n < \frac{2}{a_1} for all positive integers nn and that this is the smallest possible bound.

    Dataset text AI-MO AoPS. Quoted with source.

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