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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 $n\in \mathbb{N}$

Muradin Bronzebeard
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    Yes it is convergent. Have you any idea what the limit my be? Note that $x_0$ represents that limit in the sequel. – drhab Mar 16 '18 at 13:30
  • $(1,2)$, right? – Muradin Bronzebeard Mar 16 '18 at 13:31
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    Correct. Now you must verify that by proving that some $n^$ exists with $n>n^\implies |(1,2)-Y_n|<r$ (or equivalently $|(\frac1n,\frac2n)|<r$) where $r>0$. – drhab Mar 16 '18 at 13:34
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    You could take some $n^$ that satisfies $n^>5r^{-1}$. – drhab Mar 16 '18 at 13:39
  • Takayama does not usually show examples, so I'm having trouble finding this $n^*$ and $r$ – Muradin Bronzebeard Mar 16 '18 at 13:40
  • $r$ must not be found (it is allready there). For $n^*$ see my former comment. – drhab Mar 16 '18 at 13:41
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    $r$ is arbitrary, you are not meant to find it – Jürgen Sukumaran Mar 16 '18 at 13:41
  • in the case you mentioned $r>\frac{5}{n^*}$ – Muradin Bronzebeard Mar 16 '18 at 13:41
  • You should see this as a condition on $n^$, not on $r$. We want to show that we can find a particular value for $n^$ no matter which $r$ value you pick. Sometimes it is useful to think of convergence like a game: The opponent picks an $r$ value and then you must pick an $n^*$ to make things work. – Jürgen Sukumaran Mar 16 '18 at 13:43
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    No, I start with a fixed positive $r$ and based on that I search for some $n^$ that guarantees that $n>n^$ implies that $|(\frac1n,\frac2n)|<r$. Then I notice that $n^$ will indeed guarantee that if $n^>5r^{-1}$. Now I am finished: I found my $n^*$. I can choose e.g. for the smallest integer that exceeds $5r^{-1}$. – drhab Mar 16 '18 at 13:45

1 Answers1

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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.

Jürgen Sukumaran
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