Solve $\alpha_{n+1}=-1/\alpha_n+\beta_n$, with $\lim_{n\rightarrow\infty}\alpha_n=0$.

Question

Suppose that $(\beta_n)_{n\in\mathbb{N}}$ is a strictly positive, real-valued sequence and is known.

Can we find a $[0,\infty]$-valued sequence $(\alpha_n)_{n\in\mathbb{N}}$ satisfying the following recursion? $$\alpha_{n+1}=-1/\alpha_n+\beta_n,\quad n\in\mathbb{N};$$ with boundary condition $$\lim_{n\rightarrow\infty}\alpha_n=0. $$

I have no idea how to proceed with this one. Any help much appreciated!

Answer 1

Generally, no.

We have $\beta_n = \alpha_{n+1} + \frac1{\alpha_n}$ and with $\lim_{n\to\infty}\alpha_n = 0$ and $\alpha_n > 0$ this implies $\lim_{n\to\infty}\beta_n = \infty$.

So the given series $(\beta_n)$ converging to $\infty$ is a necessary condition for the $(\alpha_n)$ to exist.

I have no idea if it is also sufficient, I suspect not.