I want to prove that $x_n \to 0$ given the following:
- $x_n$ is a Cauchy sequence and
- for all $\epsilon > 0 $ there exists a $n > \frac{1}{\epsilon}$ (let's call this number $n_1$) such that $|x_{n_1}|<\epsilon$.
By 1), we know that for any positive $\epsilon$ there exists a $N_1$ such that $|x_n - x_m| < \epsilon $ for all $n,m \ge N_1$. My approach is to invoke the triangle inequality (specifically, $|A-B|\le |A|+|B|$) to link together 1) and 2). To do this, I want to construct a $N_2$ such that $n_1 \ge N_2 > N_1$ so that $|x_{n_1} - x_m| < \epsilon $ for any $m \ge N_2$ (since $N_2 > N_1$ any $m \ge N_2$ will satisfy the Cauchy inequality). From here we have $$|x_{n_1} - x_m| \le |x_{n_1}| + |x_m| < \epsilon + |x_m|.$$ The idea is to get rid of the $|x_{n_1}|$ in the middle and the $|x_m|$ on the right leaving just $|x_m|<\epsilon$ which would show that there exists a $N = N_2$ such that $|x_m|<\epsilon$ for all $m \ge N_2$.
I am just not sure how to do this; furthermore such a $N_2$ is not guaranteed to exist as the integers is not dense in itself. I would appreciate any hints/pointers in the right direction, especially if my reasoning is incorrect.