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.
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Sign up to join this communitySuppose $b_1, b_2, b_3, \dots \in \Bbb{R}$ 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.
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.