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Let $\left\{ x_n \right\}_{n\geq0}$ be a sequence of real numbers such that $$x_{n+1}=\lambda x_n+(1-\lambda)x_{n-1},\ n\geq 1,$$for some $0<\lambda<1$

(a) Show that $x_n=x_0+(x_1-x_0)\sum_{k=0}^{n-1}(\lambda -1)^k$

(b) Hence, or otherwise, show that $x_n$ converges and find the limit.

Note that $x_{n+1}=\lambda x_n+(1-\lambda)x_{n-1}$$$=>x_{n+1}-x_n=(\lambda-1)(x_n-x_{n-1})=\cdots=(\lambda-1)^n(x_1-x_0),\forall n$$

Hence we get $x_n-x_0=(x_n-x_{n-1})+(x_{n-1}-x_{n-2})+\cdots+(x_1-x_0)=(\lambda-1)^{n-1}(x_1-x_0)+(\lambda-1)^{n-2}(x_1-x_0)+\cdots+(x_1-x_0)=(x_1-x_0)\sum_{k=0}^{n-1}(\lambda -1)^k.$

Help me in convergence part.

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Now you have a closed form for the general term, can you see a geometric progression which you can sum? – Mark Bennet Aug 4 '12 at 16:13

3 Answers 3

up vote 1 down vote accepted

Hint: The geometric series $1+r+r^2+r^3+\cdots$ converges to $\frac{1}{1-r}$ if $|r|\lt 1$.

This is a fact that is undoubtedly already familiar to you. It can be proved by showing that $$1+r+r^2+\cdots +r^{n-1}=\frac{1-r^n}{1-r},\tag{$1$}$$ and then noting that $|r|\lt 1$, then $r^n\to 0$ as $n \to \infty$.

One way to prove $(1)$ is to show that $$(1-r)(1+r+r^2+\cdots +r^{n-1})=1-r^n.$$ This can be done by multipling out the left-hand side and observing the mass cancellation.

In our case, $r=\lambda-1$ and therefore $\frac{1}{1-r}=\frac{1}{2-\lambda}$.

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So, the limit of $x_n$ be $x_0+\frac{x_1-x_0}{2-\lambda}$ – Argha Aug 4 '12 at 16:22
@Ranabir:My previous comment was wrong. Sign error! Your $\frac{1}{2-\lambda}$ is right. – André Nicolas Aug 4 '12 at 16:54

$x_{n+1}=\lambda x_n+(1-\lambda)x_{n-1}$

If $x_n=a^n =>a=\lambda,1=>x_n=A+B\lambda^n $ where A,B are finite indeterminate constants.

Clearly, the convergence of $x_n$ depends whether $\lambda^n$ tends to a finite value or ∞ as $n→∞$. We know about infinity G.P. like $\sum_{0≤r<∞} b a^r$ which converges if |a|<1.

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What is "infinity G.P."? – Byron Schmuland Aug 4 '12 at 16:54
@ByronSchmuland My guess: Infinity geometric progression. The OP means geometric series. – Pedro Tamaroff Aug 4 '12 at 17:29
@lab: Why you consider $x_n=a^n$ and how you get $x_n=A+B \lambda ^n$ – Argha Aug 4 '12 at 19:20

Since $x_n−x_0=(x_1−x_0)\sum_{k=0}^{n-1}(\lambda-1)^k=A\frac{1-(\lambda-1)^n}{2-\lambda}$, where $A=x_1-x_0$. then $x_n=x_0+A\frac{1-(\lambda-1)^n}{2-\lambda}$. Since $0<\lambda<1$ then $(\lambda-1)^n\rightarrow 0$ as $n\rightarrow \infty$ and thus $$\lim_{n\rightarrow \infty} x_n=x_0+\frac{x_1-x_0}{2-\lambda}.$$

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