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An easy recurent sequence a_1+a_2+...+a_n<1

Source: Romanian IMO Team Selection Test TST 2003, problem 1

September 24, 2005
algebra proposedalgebra

Problem Statement

Let (an)n≥1(a_n)_{n\geq 1} be a sequence for real numbers given by a1=1/2a_1=1/2 and for each positive integer nn an+1=an2an2−an+1. a_{n+1}=\frac{a_n^2}{a_n^2-a_n+1}. Prove that for every positive integer nn we have a1+a2+⋯+an<1a_1+a_2+\cdots + a_n<1.
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