<s>[INST] Keep getting wrong answer for showing $(a_n)=(\frac{2n^2-1}{n^2},\frac{1}{n})$ is $d^{(2)}$-convergent? A sequence in $\mathbb{R}^2$ is given by:
$a_n=(\frac{2n^2-1}{n^2},\frac{1}{n})$, for each $n∈\mathbb{N}$.
I must show that $(a_n)$ is $d^{(2)}$-convergent.
My textbook says that I can show $a_n$ is convergent by showing $d^{(2)}(a_n, 0)$ converges to $0$.
So I have tried: 
$d^{(2)}((\frac{2n^2-1}{n^2},\frac{1}{n}),(0, 0))=\sqrt{(0-\frac{2n^2-1}{n^2})^2+(0-\frac{1}{n})^2}$
which when expanded ends up as:
$d^{(2)}((\frac{2n^2-1}{n^2},\frac{1}{n}),(0, 0))=\sqrt{\frac{4n^4-3n^2+1}{n^4}}$
and so:
$$\lim_{n\to\infty} \sqrt{\frac{4n^4-3n^2+1}{n^4}}=\sqrt{\lim_{n\to\infty} \frac{4n^4-3n^2+1}{n^4}}=\sqrt{\lim_{n\to\infty} {4+\frac{3}{n^2}+\frac{1}{n^4}}}=\sqrt{4}=2$$
This converges to $2$ rather than the expected $0$, so therefore it is not a real null sequence and hence $(a_n)$ is not $d^{(2)}$-convergent, which the question tells me is wrong.  Where have I gone wrong?
 [/INST] The sequence does converge, but not to $(0,0)$.  Observe that 
\begin{align*}
d^{(2)}(a_n,(2,0))&=\sqrt{\left(\frac{2n^2-1}{n^2}-2\right)^2+\left(\frac{1}{n}-0\right)^2} \\
&=\sqrt{\left(\frac{1}{n^2}\right)^2+\left(\frac{1}{n}\right)^2} \\
&=\sqrt{\frac{1+n^2}{n^4}}.
\end{align*}
Can you take it from here?
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