show that $x_n$ converges and find the limit 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.
 A: $x_{n+1}=\lambda x_n+(1-\lambda)x_{n-1}$
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.
A: 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 $\frac{1}{1-r}=\frac{1}{2-\lambda}$.
A: 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 $$\lim_{n\rightarrow \infty} x_n=x_0+\frac{x_1-x_0}{2-\lambda}.$$
