Is $\left\{Y_n\right\}=\left\{(\frac{1}{n}+1,\frac{2}{n}+2)\right\}$ is a convergent sequence?
I'm studying through Takayama - Math Economics. I have to consider an open ball $B_r(x_o)\subseteq V$ and show that $\left\{Y_n\right\}\in V$, for $n> n^*$, when $\left\{Y_n\right\}\in B_r(x_o)$, for $n>n^*$.
I can't solve this problem, and the suggested question from this page didn't help me.
p.s.: $B_r(x_o)$ is a open ball and $n\in \mathbb{N}$
It is clear that the sequence convegres to $x_0=(1,2)$, right? Now we just have to prove that this is the case. Let $B_r((1,2))$ be an open ball of radius $r$ centered at the point $(1,2)$. Then, a point $z$ is in the ball if and only if $\|z-(1,2)\|<r$. We can describe $z$ as $z=(z_1,z_2)$ and explicitly compute this difference,
$$\|z-x_0\|=\sqrt{(z_1-1)^2+(z_2-2)^2}$$
We want this difference to be less than $r$ so that $z$ is in the ball,
$$\|z-x_0\|=\sqrt{(z_1-1)^2+(z_2-2)^2}<r,\\ \implies (z_1-1)^2+(z_2-2)^2<r^2$$
Now, if $z$ is an element of the sequence at hand then we will require,
$$\left(\left(\frac{1}{n}+1\right)-1\right)^2+\left(\left(\frac{2}{n}+2\right)-2\right)^2<r^2,\\ \implies \frac{1}{n^2}+\frac{4}{n^2}<r^2$$
Now we just solve for $n$ and we are done.
$$1+4<n^2r^2,\\ \implies n^2>\frac{5}{r^2},\\ \implies n>\sqrt{\frac{5}{r^2}}$$
So for all $n>n^*=\sqrt{\frac{5}{r^2}}$, we have that $Y_n\in B_r((1,2))$ and thus the sequence $\{Y_n\}$ converges to $(1,2)$ since $r$ was arbitrary.