convergence of recursive sequence to limit not dependent on first term

Question

The question is to show that the sequence defined by $a_{n+1} = \alpha \cdot a_n + 2$ converges to a limit independent of $a_1$ if $|\alpha| < 1$. I have been able to show that the limit should be $\frac{2}{1-\alpha}$ (which does not depend on $a_1$) if it exists, but I haven't been able to show that the limit does exist. I was thinking it would be sufficient to show that the limit as $n \rightarrow \infty$ of $|a_n - \frac{2}{1- \alpha}| = 0$ (alternatively, for any $\epsilon > 0$ there's $N$ such that $n \in \mathbb{N}, n \geq N \rightarrow |a_n - \frac{2}{1- \alpha}| < \epsilon$) but I do not know how to proceed.

Answer 1

$$a_{n+1}-\frac{2}{1-\alpha}=\alpha\left(a_n-\frac{2}{1-\alpha}\right),$$ which gives that $b_n=a_n-\frac{2}{1-\alpha}$ is a geometric progression.

Id est, $$a_n-\frac{2}{1-\alpha}=\left(a_1-\frac{2}{1-\alpha}\right)\alpha^{n-1}$$ and $$\lim_{n\rightarrow+\infty}a_n=\frac{2}{1-\alpha}.$$

Answer 2

Dividing by $\alpha^{n+1}$ you have: \begin{align} \frac{a_{n+1}}{\alpha^{n+1}} &=\frac{a_n}{\alpha^n}+\frac 2{\alpha^{n+1}}\\ &=\frac{a_{n-1}}{\alpha^{n-1}}+\frac 2{\alpha^n}+\frac 2{\alpha^{n+1}}\\ &=\cdots\\ &=\frac{a_1}{\alpha}+\sum_{k=2}^{n+1}\frac 2{\alpha^k}\\ &=\frac{a_1}{\alpha}+\frac 2{\alpha^2}\sum_{h=0}^{n-1}\frac 1{\alpha^h}\\ &=\frac{a_1}{\alpha}+\frac 2{\alpha^2}\frac{1-\alpha^{-n}}{1-\alpha^{-1}} \end{align} from which \begin{align} a_{n+1} &=a_1\alpha^n+2\frac{\alpha^n-1}{\alpha-1}\\ &\to\frac 2{1-\alpha} \end{align} because $\alpha^n\to 0$ for $0<\alpha<1$.