Suppose $b\_1, b\_2, b\_3, \dots \in \Bbb{R}$ satisfy the Riccati-type recurrence
$$b\_{k+1}=\frac{1+kb\_k}{k-b\_k},\quad k\ge 1.$$



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> Is it true that such a sequence reaches infinitely many positive as well as negative values? It appears to be so.
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Riccati-type recurrence: infinitely many sign changes?
Yes, choose $a\_1$ such that $\frac{1}{b\_1}=\tan(a\_1)$, and let $a\_{k+1}=a\_k-\arctan\frac{1}{k}$, then we have $\frac{1}{b\_k}=\tan(a\_k)$, since $\sum\_k \arctan\frac{1}{k}=\infty$, the conjecture follows.


