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?
 A: 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.
A: Let's prove first that $(x_n)$ must be bounded. Indeed, there is some $N\in \mathbb N$ such that $$n\geq N\implies |3x_{n+1}-x_n-1|\leq 1$$ 
Since $3|x_{n+1}|-|x_n+1|\leq |3x_{n+1}-x_n-1|$, we get $|x_{n+1}|\leq \frac{2+|x_n|}3$. 
Let $M=\max(1,|x_N|)$. 
Note that $|x_N|\leq M$ and if $|x_n|\leq M$ then $|x_{n+1}|\leq \frac{2+M}3\leq M$ (the last inequaliy holds since $M\geq 1$). Hence $\forall n\geq N, |x_n|\leq M$ and $(x_n)$ is bounded by $\max(|x_0|,\ldots,|x_{N-1}|,M)$.
Since $(x_n)$ is bounded, it has a convergent subsequence $x_{n_k}\to x$. 
But since $3x_n - x_{n-1} \to 1$, we have $3x_{n_k}-x_{n_k-1} \to x$ hence $x_{n_k-1}\to 3x-1$. 
Iterating this, $x_n$ has subsequences with arbitrarily large limits, hence $x_n$ is unbounded, a contradiction.
Thus, $x_n$ is bounded and has a unique accumulation point. A classic lemma yields convergence of $x_n$ to this accumulation point.
