Let $x_0=1$ and
$$x_{k+1} = (1-a_k)\left(\frac{3}{2}-\frac{1}{2}\frac{1}{x_k}\right)$$
where $a_n$ is a known sequence satisfying that $a_k\in(0,1)$ for all $k$ and $a_k\to 0$ as $k\to\infty$. How to prove that $x_k\to 1$ as $k\to\infty$?
The difficulty here is that
- It is not known how fast $a_k$ converges to zero, and I don't know how it affect the convergence of $x_k$;
- $x_k$ may change sign and is not monotone, so I don't know how to prove $x_k$ even converges;
- Furthermore, if we assume $x_k$ do converge to some limit $x^*$, then by taking the limit,
$$x^*=(1-0)\left(\frac{3}{2}-\frac{1}{2}\frac{1}{x^*}\right)$$
I find there are two possible solution $x^*=1/2$ or $x^*=1$. How to exclude the case that $x^*=1/2$?
Best Answer
If, as you say, $a_k<0.1$ for all $k$, then we can prove by induction that $x_k>\frac{3}{4}$ for all $k$, with induction step $x_{k+1}>0.9\left(\frac{3}{2}-\frac{1}{2\cdot\frac{3}{4}}\right)=\frac{3}{4}$. By a similar induction we get $x_k\in\big[\frac{3}{4},1\big]\;\forall k$.
So $1-x_k\in\big[0,\frac{1}{4}\big]$ for all $k$. Now note that $$ \begin{split} 1-x_{k+1} &= 1-(1-a_k)\left(\frac{3}{2}-\frac{1}{2x_k}\right)\\ &=-\frac{1}{2}+\frac{1}{2x_k}+a_k\left(\frac{3}{2}-\frac{1}{2}x_k\right)\\ &\leq\frac{1-x_k}{2x_k}+3a_k\leq\frac{2}{3}(1-x_k)+3a_k. \end{split} $$
So as $a_k\to0$, we also have $1-x_k\to 0$.