Cauchy sequence doesn't converge to zero has all the same sign for values after a certain point

Question

Show that if a Cauchy sequence does not converge to 0, all the terms of the sequence eventually have the same sign.?

Answer 1

What do we know? We know that $a_{k}$ is a Cauchy sequence, which means for every $\epsilon > 0$, we can find some point $N$ in the sequence such that for all points after $N$, they are all less than $\epsilon$ away from each other, i.e., for all $n, m \geq N$, $|a_{n} - a_{m}| < \epsilon$, right?

Now, we know by assumption $a_{k} \not \to 0$, which means there is some $\epsilon > 0$ such that for every point $N$ in the sequence, we can find a later point not within $\epsilon$ of $0$, i.e., $\exists n > N$ with $a_{n} \not \in (0-\epsilon, 0 + \epsilon)$.

But for this very epsilon, by the Cauchy-ness of $a_{k}$, we know that there is a point $N'$ such that all later points are within $\epsilon$ of each other. In other words, all later points after $N'$ must either be $< -\epsilon$ or $>\epsilon$, since otherwise there would be later points past $N'$ that are more than $\epsilon$ away from each other -- which contradicts the Cauchy-ness of $a_{k}$.

Thus, since all later points past $N'$ are either $< -\epsilon$ or $> \epsilon$, that means all later points are one sign, either all negative or all positive.

Answer 2

$(x_{n})$ is Cauchy implies that it converges. Suppose that it converges to $L>0$. There exist an $N$ such that for all $n>N$ we have $| x_{n} - L|<L$. So for all $n>N$, $x_{n}>0$. A similar proof works for $L<0$.