Convergence of a sequence

Question

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

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 $x^*=1/2$?

Answer 1

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 $1-x_k\to 0$.

Answer 2

Taking $b_k=(1-a_k), y_k=\sqrt{2}x_k, c=\frac{3}{\sqrt{2}}$ we have $y_{k+1}=b_k(c-\frac{1}{y_k})$. Now as $b_k \rightarrow 1$ assuming it's variation to be sufficiently small for large $k$, we take $b_k=1 \forall k\geq k_0$.

The recurrence relation becomes $y_{k+1}=c-\frac{1}{y_k},k≥k_0$. The constancy points of this recurrence is at $\frac{1}{\sqrt{2}}$ and $\sqrt{2}$. So, we have $\Delta y=-\frac{(y-1/\sqrt{2})(y-\sqrt{2})}{y}$,which is positive when $y \in [1/\sqrt{2},\sqrt{2}]$, with the maximum being at $1$. Now, for some $y_k$ in this interval ($I$); $(\sqrt{2}-y)-\Delta y_k=\frac{1-y/\sqrt{2}}{y} \geq 0$ implying that $y_{k+1} \in I; y_{k+1}>y_k$.

So, if $y_{k_0} \in I$, the sequence converges to $\sqrt{2}$ from left. While if $y_{k}>\sqrt{2}$, $y_{k+1}=\frac{3}{\sqrt{2}}-\frac{1}{y}\in (\sqrt{2}, \frac{3}{\sqrt{2}})$, hence $y_k$ converges to $\sqrt{2}$ from right.

Similarly for $y_{k}<0$, $y_{k+1}=\frac{3}{\sqrt{2}}+\frac{1}{|y_k|} >\sqrt{2}$. The secuence converges.

Lastly, when $0<y<\frac{1}{\sqrt{2}}, \Delta y_k<0$ and decreasing very fast towards zero. So, $y_{k+r}<0$ for some $r>0$ which then converges to $\sqrt{2}$ as previously mentioned. So, the sequence $x_k$ converges to $1$.