TITLE: Riccati-type recurrence: infinitely many sign changes?
QUESTION [9 upvotes]: Suppose $b_1, b_2, b_3, \dots \in \Bbb{R}$ satisfy the Riccati-type recurrence
$$b_{k+1}=\frac{1+kb_k}{k-b_k},\quad k\ge 1.$$

Is it true that such a sequence reaches infinitely many positive as well as negative values? It appears to be so.

REPLY [8 votes]: 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.