# Limit of sequence $x_n$

For $$\alpha \in (0,1)$$ let the sequence $$\{x_n\}$$ be such that $$x_0 = 0, x_1 = 1$$ and $$x_{n+1} = \alpha x_n + (1-\alpha)x_{n-1},\quad n\geq1$$. Find $$\displaystyle \lim_{n\rightarrow \infty}x_n$$.

My try:

Since $$\alpha \in (0,1)\implies (\alpha-1)\in (-1,0)$$

\begin{align}|x_{n+1}-x_n|=&|\alpha -1||x_n-x_{n-1}|\\ &\vdots\\ &|\alpha -1|^n|x_1-x_0|\end{align} \implies \{x_n\} \rightarrow 0

Is there any alternative proof ?

• ${x_n}\to0$ not true. – Samvel Safaryan Nov 12 '18 at 8:06
• Can you please suggest what is wrong in my proof? – Yadati Kiran Nov 12 '18 at 8:07
• Oh! I guess I have shown that the sequence is convergent but not the limit. – Yadati Kiran Nov 12 '18 at 8:09
• $|x_{n+1}-x_n|=|\alpha-1|^n$, but it doesn't follow that $lim_{n\to+\infty}{x_n}=0$. – Samvel Safaryan Nov 12 '18 at 8:13
• Below is my solution, you can prove by induction that $x_n$ satisfies all the conditions of the problem. – Samvel Safaryan Nov 12 '18 at 8:17

Let's look for $$x_n$$ in the form $$\lambda^n$$ $$\\\lambda^2=\alpha\lambda+(1-\alpha) \\\lambda^2-1-\alpha(\lambda-1)=0 \\(\lambda-1)(\lambda+1-\alpha)=0$$ $$=>\lambda=1,\;\alpha-1=>x_n=A+B(\alpha-1)^n\\ 0=x_0=A+B\\ 1=x_1=A+B(\alpha-1) \\B(\alpha-2)=1=>B=\frac{1}{\alpha-2}=>x_n=\frac{1}{2-\alpha}\Big(1-(\alpha-1)^n\Big)=>\lim_{n\to+\infty}{x_n}=\frac{1}{2-\alpha}$$