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For $\alpha\in(0,1)$, let $(x_{n})$ be a sequence such that

$$x_{n}=\begin{cases} 0, \ n=0 \\ 1, \ n=1 \\ \alpha x_{n-1}+(1-\alpha) x_{n-2},\ n\ge 2 \end{cases}$$

Problem. Find $\lim_{n\rightarrow \infty}x_n$.


For $n\ge2$ we have

\begin{alignat}{3} x_2 &= \alpha +( 1-\alpha)\cdot 0 &&=\alpha\\ x_3 &= \alpha x_2 +(1-\alpha)x_1 &&=\alpha^2 - \alpha +1 \\ x_4 &= \alpha x_3 +(1-\alpha)x_2 &&=\alpha^{3}-2\alpha^2 +2\alpha\\ \end{alignat} $$\vdots$$ $$x_{n+1}=\alpha x_n + (1- \alpha) x_{n-1}$$

But I don't know how to approach the limit.

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  • $\begingroup$ @AlexD I almost rolled back your edit, but settled in the end to only fix the (critical) typo in the definition of the recurrence relation itself. Please be more considerate before making such massive edits in the future. The point of a good edit is to render OP's question more clear and more readable. It is not to completely rewrite the question as you would have asked it, instead. $\endgroup$ – dxiv May 1 '18 at 5:51
  • $\begingroup$ @dxiv Thank you, I'll consider this before making an edit in the future. My apologies. $\endgroup$ – Alex D May 1 '18 at 6:02
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Hint: $\;x_{n+1}\color{red}{-x_n} =\alpha x_n\color{red}{-x_n} + (1- \alpha) x_{n-1} \iff x_{n+1}-x_n =(\alpha-1)(x_n-x_{n-1})\,$, so $\,x_{n+1}-x_n\,$ is a geometric progression, and therefore $\,x_n\,$ is the sum of a geometric progression.


[ EDIT ]   Followup hint: $\;x_{n+1}-x_n =(\alpha-1)(x_n-x_{n-1})$ $=(\alpha-1)^2(x_{n-1}-x_{n-2})$ $=\ldots$ $=(\alpha-1)^n(x_1-x_{0})=(\alpha-1)^n\,$.

Then $\,x_n = (x_n-x_{n-1}) + \ldots+(x_1-x_0)+x_0=(\alpha-1)^{n-1}+(\alpha-1)^{n-2}+\ldots+1=\ldots\,$

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  • $\begingroup$ that mean $lim_{n\rightarrow \infty}x_n = lim_ {n\rightarrow \infty}1+( α-1) + (α-1)^2 +.....+(α-1)^n =lim_{n\rightarrow \infty}\frac{1({1-(α-1)^n})}{1-(α-1)} =\frac{1}{2-\alpha}$.....Am i right ? or Am i wrong @dxiv Pliz tell me $\endgroup$ – user557742 May 1 '18 at 5:27
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    $\begingroup$ @Henry The limit came out right, but your $\,x_n\,$ was off-by-one, see my edit. $\endgroup$ – dxiv May 1 '18 at 5:28
  • $\begingroup$ thanks a lots@dxiv $\endgroup$ – user557742 May 1 '18 at 5:29
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HINT: Prove that $x_n=\frac{(\alpha-1)^n-1}{\alpha-2}$ inductively

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  • $\begingroup$ im not getting how did u get $x_n=\frac{(\alpha-1)^n-1}{\alpha-2}$ $\endgroup$ – user557742 May 1 '18 at 4:53
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    $\begingroup$ technically, u can compute the formula of a sequence using matrix. See this post comeoncodeon.wordpress.com/2011/05/08/… $\endgroup$ – chí trung châu May 1 '18 at 4:56

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