$a=0,\, b=0$ don't satisfy the given condition and $a=0,\, b \neq 0$ satisfy it. Let $a \neq 0$, denote $c = \frac{b}{a}$. We have: (1) $ x_{n+1} = c x_n + \epsilon_n $, where $\epsilon_n \to 0$, as $n \to \infty$. From (1) we get: (2) $ x_{n+1} = \sum_{j=0}^{n-1} c^j \epsilon_{n-j} \, + c^n x_1 $. If $c \ge 1$ and if we put $ \epsilon_j = \frac{1}{j}, \, x_1=0 $ we will get $x_{n+1} > \sum_{j=1}^n \frac{1}{j}$, so $ x_n \to \infty$. The case $ c \le -1$ is similar. Let now $|c| < 1$, we will prove that $ x_n \to 0$. Because $\epsilon_n$ is convergent $ \Rightarrow \, \forall n \, |\epsilon_n| < A $, for some $A>0$. Let fix some $ \epsilon >0$. Then there exists $N$, such that $ |\epsilon_j| < \epsilon $ when $ j >N $. Let estimate $x_n$ when $n > N$. From (2) we get: $ |x_{n+1}| < \sum_{j=0}^{n-N} |c|^j |\epsilon_{n-j}| + \sum_{n-N+1}^{n-1} |c|^j |\epsilon_{n-j}| \, +|c|^n |x_1| <$ $< \epsilon \sum_{j=0}^{n-N} |c|^j \, + \, A \sum_{n-N+1}^{n-1} |c|^j \, + |c|^n |x_1|< $ $ < const(c) \epsilon + const(c) |c|^{n-N} A + |c|^n |x_1| $. (where $const(c)$ depends only on $c$ ) Now, because $|c| < 1$, it is obvious from the last expression that when $n$ is large enough $|x_n| < 2 const(c) \epsilon$, which means $ lim_{n \to \infty} x_n = 0$. Finnaly: all $(a,b)$ are: 1. $a=0, \, b \neq 0 $ 2. $a \neq 0, \, |\frac{b}{a}| < 1 $.