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 $k\to\infty$?




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The difficulty here is that


1. It is not known how fast $a\_k$ converges to zero, and I don't know how it affect the convergence of $x\_k$;
2. $x\_k$ may change sign and is not monotone, so I don't know how to prove $x\_k$ even converges;
3. 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 $x^\*=1/2$?



Convergence of a sequence
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 $1-x\_k\to 0$.


