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 $|x_n-x_m| < \frac{1}{n} + \frac{1}{m}+ \epsilon< 3 \epsilon$