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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?

1 Answers 1

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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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    From what you've given me I make the limit to be 0 and hence $d^{(2)}(a_n, (2,0))$ is a null sequence as I needed. Have I misunderstood in thinking that I should be checking whether it converges to $(0, 0)$ in order to work out if it is a real null sequence? If this is the case, how should I decide to check $(2, 0)$? Thanks for your response. – 2017-02-21
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    I've just reread the definition and it says a metric d-converges to $a$,if $d(a_n, a)$ is a null sequence. So as you noted, I put in the wrong value of a. Should I have checked the limits of $(a_n)$ first, and then put that value in for a instead? – 2017-02-21
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    @MikeA Yes, with such problems it is a good idea to check what the individual components converge to (i.e. $\frac{2n^2-1}{n^2}\to 2$ and $\frac{1}{n}\to0$) to find a candidate for the limit of the sequence $(a_n)$. **SPOILER ALERT** Someday soon, you will be able to show that $(a_n)$ converges to $a$ if and only if each component sequence of $(a_n)$ converges to the corresponding component of $a$. – 2017-02-21
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    Ha, love the spoiler alert, should be interesting when I come to it. Anyhow, thanks for your help it all makes sense now! – 2017-02-21