Let $x_0=x_1=2$ be the initial condition. Consider the following recursion: $$x_{k+1}^{-2}=((1+\alpha)x_k - \alpha x_{k-1})^{-2}+\beta$$ where $\alpha\in(0,1)$ and $\beta>0$. Prove that
- Given an $\alpha$, there exists a small enough $\beta$ such that $x_k$ is decreasing and $\lim_{k\to\infty} x_k=0$.
- Under the assumption of part 1, we have $x_k=\Theta(\frac{1}{\sqrt{k}})$.
To simplify the problem, we can consider a special case, i.e., $\alpha=\beta=1/2$. In this case, I ran some numerical experiments and the results show that $x_k$ is indeed decreasing with $\lim_{k\to\infty} x_k=0$. Furthermore, $\lim_{k\to\infty}kx_k^2=1$. Can someone give a hint at least on the special case? I have tried to use Taylor expansion on the term $((1+\alpha)x_k - \alpha x_{k-1})^{-2}$, but since both $x_k$ and $x_{k-1}$ converge to $0$, I don't know how shall I expand the series.