Suppose $(x_n)$ is a convergent sequence and $(y_n)$ satisfies the following condition:
$\forall \epsilon>0, \exists\ m\in \mathbb N:|x_n-y_n|<\epsilon\ \forall\ n\geq m$
Prove that the sequence $(y_n)$ is convergent.
I KNOW THAT THIS QUESTION HAS BEEN ASKED BEFORE HERE but I want to verify my method which again is a bit long.
My approach:
Let $\lim(x_n)=L$, therefore $\forall\ \epsilon>0, \exists\ k\in \mathbb N:|x_n-L|<\epsilon\ \forall\ n\geq k$
$$\text{We have}\ \forall\ \epsilon>0, \exists\ m\in \mathbb N:|x_n-y_n|<\epsilon\ \forall\ n\geq m$$
Let $\epsilon = |x_n-L|$ and $r=\sup\{m,k\}$.
$$\therefore |x_n-y_n|<|x_n-L|\space\forall\ n\geq r$$
Thus $(x_n-y_n)^2<(x_n-L)^2$ $$y_n^2-2x_ny_n<L^2-2x_nL$$ $$(y_n-L)(y_n+L)<2x_n(y_n-L)$$ Suppose $y_n<L\space \forall\ n\geq r$, this implies that $(y_n)$ must be convergent.
Suppose $y_n>L \space \forall\ n\geq r$, then $y_n<y_n+L<2x_n\space \forall\ n\geq r$
Now since $x_n$ is convergent, therefore $(y_n)$ must be convergent.
If $y_n=L$, then it would yield in a constant sequence which again, is convergent.
$\therefore $ The sequence $(y_n)$ is convergent in all cases.
Please help me verify this method. Any correction would be highly appreciated
Thanks