Does $x_n$ converge, given $\lim(3x_{n+1} - x_{n})=1 $ I want to prove that $x_n$ converges, given that $\lim (3x_{n+1} - x_n ) = 1$
Attempt: Since $\lim (3x_{n+1} - x_n ) = 1$, set $\epsilon > 0, $ such that $\forall n > N, |3x_{n+1} - x_n -1 | < \epsilon. $ Then, 
$$|x_{n+1} - x_n|< \min(\frac{\epsilon + 1 - 2x_N}{3},\frac{1 -\epsilon - 2x_N}{3},1)\ .$$ 
So 
$$|x_{n+k} - x_n| \leq |x_{n+k} - x_{n+k-1}| + |x_{n+k-1} - x_{n+k-2}| + ... + |x_{n+1}-x_n|< $$
$$< k \times \min(\frac{\epsilon + 1 - 2x_N}{3},\frac{1 -\epsilon - 2x_N}{3},1)$$
I tried showing that $x_n$ is Cauchy but there seems to be some problems, any advice?

The limit, if it exists, is $\frac{1}{2}$, so I think it is easier to work with $u_n:=x_n-\frac{1}{2}$. The hypothesis then is $3 u_{n+1} - u_n \to 0$. We want to prove that $u_n\to 0$.
Let $\epsilon>0$. For big enough $n$ we can get $3 |u_{n+1}|\leqslant |u_n| +\epsilon$, and then 
$$
|u_{n+k}|\leqslant \frac{1}{3^k}|u_n| + \frac{\epsilon}{3}.\frac{1}{1-\frac{1}{3}}$$
which will  give the result.
