find $\lim_{n\rightarrow \infty}x_n$ ?.

Question

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

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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.

Answer 1

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.

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

Answer 2

HINT: Prove that $x_n=\frac{(\alpha-1)^n-1}{\alpha-2}$ inductively