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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 $x^*=1/2$?
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    $\begingroup$ A more basic problem is that $x_k$ need not even be defined for all $k$ (when $x_{k-1}=0$). $\endgroup$ Commented Oct 11, 2022 at 21:48
  • $\begingroup$ Even if $x_k$ is defined for all $k$, you can find $a_0$ so that $x_1$ is slightly less than $\frac{1}{2}$ and then adjust $a_k$ so that $x_k=\frac{1}{2}(x_{k-1}+\frac{1}{2})$, so in general this does not converge to $1$ $\endgroup$
    – Saúl RM
    Commented Oct 11, 2022 at 21:58
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    $\begingroup$ Thanks for your reply. In this case what additional assumption should be made on $a_k$? Like if I know $a_k$ is always very small, e.g., $a_k<0.1$? $\endgroup$ Commented Oct 11, 2022 at 22:16

2 Answers 2

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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 $1-x_k\to 0$.

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    $\begingroup$ smart argument, thanks! $\endgroup$ Commented Oct 12, 2022 at 2:31
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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 $1$.

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  • $\begingroup$ Thanks for your reply. Intuitively I understand why you can take $b_k=1$. However, I am not sure how to argue rigorously that if you prove the case for $b_k=1$, then it also implies the same result for $b_k\to 1$. Can you explain more? How do you avoid the case Saúl RM mentioned in the comment? $\endgroup$ Commented Oct 12, 2022 at 13:52
  • $\begingroup$ @Jean Legall, thanks for pointing this out. I'll think about it. $\endgroup$
    – Alapan Das
    Commented Oct 12, 2022 at 15:53

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