If $(x_n)$ is a convergent sequence, for any $\epsilon>0, \exists M$ such that $|x_n-y_n|<\epsilon$ for all $n \geq M$. Is $(y_n)$ convergent?

Question

If $(x_n)$ is a convergent sequence and $(y_n)$ is such that for any $\epsilon>0, \exists M$ such that $|x_n-y_n|<\epsilon, \forall n \geq M$. Is $(y_n)$ convergent?

My attempt:

Let lim $(x_n)=x$, then $\exists k_1 \in \mathbb N$ such that $|x_n-x|<\epsilon, \forall n\geq k_1$

Let $k=$sup {$k_1,M$}

Then $\forall n\geq k, |y_n-x|=|y_n-x_n+x_n-x|$

$\leq |x_n-y_n|+|x_n-x|<\epsilon+\epsilon=2\epsilon$

Hence $(y_n)$ converges to $x$. Is this proof correct?

Answer 1

Yes. your proof is true. But better to take $\frac{\epsilon}{2}$ in your proof.