# Cauchy with terms of differing signs after some term $\implies$ convergence

Let $(x_n)_{n=1}^{\infty}$ be Cauchy.

For each natural number $N$ (natural numbers starting at $1$), suppose there is an $n_1 \geq N$ and $n_2 \geq N$ such that $x_{n_1} < 0$ and $x_{n_2} > 0$. Using the definitions of Cauchy sequences and convergence of a sequence, show that $(x_n)_{n=1}^{\infty}$ converges to $0$.

Attempt (this is basically scratch work though):

Let $\epsilon > 0$ and $N \in \mathbb{N}$, there is a $k \geq N$ and $n \geq N$ such that $x_k < 0$ and $x_n > 0$. Thus, for all $\epsilon > 0$ and $n^{\prime} \geq N$, $$|x_{n^{\prime}} - 0| = |x_{n^{\prime}} - x_k + x_k - x_n + x_n| \leq |x_{n^{\prime}} - x_k|+|x_{k} - x_n|+|x_n|$$ First two terms are good since we have a Cauchy sequence here. How do I handle the last term?

Hints are appreciated; I don't want a complete solution.

• You probably meant "Let $\varepsilon>0$ and $N\in\mathbb{N}$" (note the upper case $N$). – Micapps Jun 15 '16 at 13:19
• @Micapps Yes, I did. Thanks. – Clarinetist Jun 15 '16 at 13:19
• Note that since $x_n > 0 > x_k$ you have $|x_n| < |x_n - x_k|$. This should help with the last term. – Micapps Jun 15 '16 at 13:22

Since $\{x_n\}$ is Cauchy it is also convergent. You can then use the fact that convergent subsequences of a convergent sequence converge to the same limit and consider the subsequence of all positive or all negative terms.