Baltic Way 1990 · Problem 3

Algebra

Let $a_{0}>0, c>0$ and

a_{n+1}=\frac{a_{n}+c}{1-a_{n} c}, \quad n=0,1, \ldots

Is it possible that the first 1990 terms $a_{0}, a_{1}, \ldots, a_{1989}$ are all positive but $a_{1990}<0$ ?

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Review

Topics

Sequences and recurrences

Solutions

Solution

Solution: Obviously we can find angles $0 < \alpha, \beta < 90^{\circ}$ such that $\tan \alpha > 0$ , $\tan (\alpha + \beta) > 0$ , $\ldots$ , $\tan (\alpha + 1989 \beta) > 0$ but $\tan (\alpha + 1990 \beta) < 0$ . Now it suffices to note that if we take $a_{0} = \tan \alpha$ and $c = \tan \beta$ then $a_{n} = \tan (\alpha + n \beta)$ .

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AoPS · Baltic Way 1990