Suppose for all positive integers $n$, $|x_n-y_n|< \frac{1}{n}$ Prove that $(x_n)$ is also Cauchy.

Question

Let $(x_n)$ be a sequence and $(y_n)$ be a Cauchy sequence. Suppose for all positive integers $n$, $$|x_n-y_n|< \frac{1}{n}$$ Prove that $(x_n)$ is also Cauchy.

My attempt: Suppose that $n \geq m$. Notice that $|x_n-x_m| \leq |x_n-y_n| + |y_n -y_m|+ |y_m-x_m|$

Since $(y_n)$ is a Cauchy sequence, for all $\epsilon >0$ , there exists $N \in \mathbb{N}$ such that for all $n \geq m \geq N$, $|y_n-y_m|< \epsilon$

Hence, we have $|x_n-x_m| \leq |x_n-y_n| + |y_n -y_m|+ |y_m-x_m| < \frac{1}{n} + \frac{1}{m}+ \epsilon =\frac{m+n}{nm}+ \epsilon < m+n+ \epsilon < \epsilon$ since $m,n \in \mathbb{N}$

Hence, $(x_n)$ is a Cauchy.

Is my proof valid?

EDIT: Choose $N$ such that $N=\frac{1}{\epsilon}$. Hence, for all $n \geq m \geq \frac{1}{\epsilon}$, we have $|x_n-x_m| < \frac{1}{n} + \frac{1}{m}+ \epsilon< 3 \epsilon$

Answer 1

Not quite, you have $n+m + \epsilon < \epsilon$, which cannot hold if $m,n \ge 0$.

Choose $N$ large enough so that $\frac{1}{N} < \epsilon$. Then, if $m,n \ge N$, you have $|x_n-x_m| \leq |x_n-y_n| + |y_n -y_m|+ |y_m-x_m| < \frac{1}{n} + \frac{1}{m}+ \epsilon < 3 \epsilon$.

Answer 2

Hint. Apply the Archimedean Principle for $\frac{1}{n}$ and $\frac{1}{m}$.

Answer 3

As $(y_n)$ is cauchy so it converges(let $\lim_{n\to \infty}y_n=y$) and the above inequality implies that $(x_n)$ also converges (that also to $y$). Now as any convergent sequence is cauchy so $(x_n)$ must also be cauchy.