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If $x_1$, $x_2$ are arbitary real numbers, $\alpha\in (0,1]$ and $x_n=\alpha x_{n-1}+(1-\alpha)x_{n-2}$ for every positive integer $n$ (>2), show that the sequence $\{x_n\}$ is convergent. (Given $x_1<x_2$)

I cannot rearrange to test convergency. Please help me by solving the problem.

Edit

I want to prove by using the property below:

If $\{x_{2n}\}$ and $\{x_{2n-1}\}$ converges to same limit $l$ then $\{x_{n}\}$ converges to $l$.

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    $\begingroup$ Hint: consider $x_n-x_{n-1}$ $\endgroup$ – Mark Bennet Dec 18 '16 at 17:40
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Note that $x_{n+2}\in[x_n,x_{n+1}]$ (or $[x_{n+1},x_n]$). Note also that the length of this interval is $$\ell_n=|x_{n+1}-x_n|=|\alpha x_n+(1-\alpha)x_{n-1}-x_n|=(1-\alpha)\ell_{n-1}$$

Since $\ell_n\to 0$, apply nested intervals.

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  • $\begingroup$ how to use nested intervals and what will be the result $\endgroup$ – user1942348 Dec 18 '16 at 19:04
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We have $x_n-x_{n-1}=(\alpha-1)(x_{n-1}-x_{n-2})$ so we have $x_n-x_{n-1}=(\alpha-1)^{n-2}(x_2-x_1)$ and hence $$\begin{aligned}x_n-x_{n-1}&=(\alpha-1)^{n-2}(x_2-x_1)\\\\x_{n-1}-x_{n-2}&=(\alpha-1)^{n-3}(x_2-x_1)\\&\vdots\\x_2-x_1&=(\alpha-1)^0(x_2-x_1)\end{aligned}$$ and summing those we get $x_n-x_1=(x_2-x_1)\frac{1-(\alpha-1)^{n-1}}{2-\alpha}$ which implies that $$\color{red}{\lim\limits_{n\to\infty}x_n=\dfrac{x_2+(1-\alpha)x_1}{2-\alpha}}.$$

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    $\begingroup$ Please check the denominator in $\frac{1-(\alpha-1)^{n-1}}{1-\alpha}$ after summing $\endgroup$ – user1942348 Dec 18 '16 at 19:14
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    $\begingroup$ Fixed. Thank you for the observation @user1942348 $\endgroup$ – CIJ Dec 18 '16 at 19:20
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$$x_n-x_{n-1}=(-1)(1-\alpha)({x_{n-1}-x_{n-2})}\\=(-1)^2(1-\alpha)^2(x_{n-2}-x_{n-1})\\=\vdots\\=(-1)^n(1-\alpha)^n(x_1-x_0)$$

Now you can complete

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  • $\begingroup$ Thanks. I understand. $x_n-x_{n-1}$ depends on $n=$even or odd. How to conclude convergency. Please help. $\endgroup$ – user1942348 Dec 18 '16 at 18:02
  • $\begingroup$ @user1942348 $(1-\alpha)<1\implies (1-\alpha)^n\to ?$ $\endgroup$ – Qwerty Dec 18 '16 at 18:11
  • $\begingroup$ It goes to zero. But how to check the subsequences $x_{2n}$ and $x_{2n-1}$ converges to same limit? How to check they are increasing/decreasing and bounded above/below? $\endgroup$ – user1942348 Dec 18 '16 at 18:15
  • $\begingroup$ @user1942348 You are forgetting the $||$ - the mod value ! $\endgroup$ – Qwerty Dec 18 '16 at 18:17
  • $\begingroup$ I think you are telling about cauchy ciretia. I want to use the subsequencial criteria to prove it. $\endgroup$ – user1942348 Dec 18 '16 at 18:21

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