Valentin Vornicu 25.09.2005 01:44 Let $(a_n)_{n\geq 1}$ be a sequence for real numbers given by $a_1=1/2$ and for each positive integer $n$ \[ a_{n+1}=\frac{a_n^2}{a_n^2-a_n+1}. \] Prove that for every positive integer $n$ we have $a_1+a_2+\cdots + a_n<1$.
Arne 25.09.2005 12:27 Define $b_n = \frac{1}{a_n}$. Then $b_{n + 1} = b_n^2 - b_n + 1$ or \[\frac{b_{n + 1} - 1}{b_n - 1} = b_n.\] Multiplying these relations (the whole thing telescopes) we get \[b_{n + 1} = b_n\cdots b_2b_1 + 1 \Leftrightarrow \frac{1}{b_n} = \frac{1}{b_1b_2\cdots b_{n - 1}} - \frac{1}{b_1b_2\cdots b_{n}}.\] Telescoping again we get \[\frac{1}{b_1} + \frac{1}{b_2} + \cdots + \frac{1}{b_n} = 1 - \frac{1}{b_1b_2\cdots b_n}\] and we are done.