# Convergence of a sequence

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 $$x^*=1/2$$?
• A more basic problem is that $x_k$ need not even be defined for all $k$ (when $x_{k-1}=0$). Commented Oct 11, 2022 at 21:48
• 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 $1$ Commented Oct 11, 2022 at 21:58
• 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., $a_k<0.1$? Commented Oct 11, 2022 at 22:16

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 $$1-x_k\to 0$$.

• smart argument, thanks! Commented Oct 12, 2022 at 2:31

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 $$0 and decreasing very fast towards zero. So, $$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$$.

• 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? Commented Oct 12, 2022 at 13:52
• @Jean Legall, thanks for pointing this out. I'll think about it. Commented Oct 12, 2022 at 15:53