# 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?

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?

• yes your proof is correct..... – Urban PENDU May 2 '15 at 18:51
• Looks fine. You could also say $y_n = x_n +(y_n-x_n).$ The hypotheses say exactly that $(y_n-x_n) \to 0.$ Therefore $\lim y_n = \lim x_n.$ – zhw. May 2 '15 at 18:53

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