Determine whether the sequence converge or diverge The sequence is 

$$a_{n+1} = 2a_n - 1$$

and 

$$a_1 = 1$$

The sequence should be 

$$\{1, 1, 1, 1, 1, 1, ...\}$$

and I get
$$\lim_{n\to\infty}a_n = 1$$

However
$$\lim_{n\to\infty}s_n = s = \infty$$

So is the sequence convergent or divergent? In other words, how do I judge the sequence?
$\displaystyle a_{n + 1} - 1 = 2\left(a_{n} - 1\right) \implies
a_{n + 1} = 2^{n}\left(a_{1} - 1\right) = 0\quad\mbox{because}\ a_{1} = 1$.


  In general, the sequence diverges whenever $\displaystyle a_{1} \not= 1$.
If the seq. is convergent, then $\lim_n a_{n+1}=\lim_n a_{n}$. Now for this limit, let $x=\lim a_{n+1}=\lim a_n$, then

$$
x=2x-1
$$
has solution $x=1$. Can you now prove that for every $\epsilon&gt;0, \exists n \in \mathbb{N}$ s.t $\forall m&gt;n$ we have $|a_m-1|&lt;\epsilon$?