MathDB

4

Part of 2002 German National Olympiad

Problems(1)

Inequality with sequences

Source: Germany 2002 - Problem 4

12/15/2022
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.
inequalitiesConvergencealgebraSequencerecursive