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.