- #1

quasar987

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[tex]E = \left\{ \frac{n^2 + 3n + 5}{n^2 + 2} \vert n \in \mathbb{N} \right\} [/tex]

Now that is easy, we first rearange the terms so we see what happens in this mess when n varies. I did the following thing..

[tex]E = \left\{ \frac{n^2 + 2 + 3n + 3}{n^2 + 2} | n \in \mathbb{N} \right\} [/tex]

[tex]E = \left\{ \frac{n^2 + 2}{n^2 + 2} + \frac{3n + 3}{n^2 + 2} | n \in \mathbb{N} \right\} [/tex]

[tex]E = \left\{ 1 + \frac{3n}{n^2 + 2} + \frac {3}{n^2 + 2} | n \in \mathbb{N} \right\} [/tex]

[tex]E = \left\{ 1 + \frac{3}{n + \frac{2}{n}} + \frac {3}{n^2 + 2} | n \in \mathbb{N} \right\} [/tex]

Now it is clear that 1 is the only accumulation point and it happens when n is arbritrarily humongeous. But we must prove it! Usually, for sets that look like

[tex]E = \left\{ \frac{1}{n} | n \in \mathbb{N} \right\} [/tex]

or

[tex]E = \left\{ \frac{1}{2^n} | n \in \mathbb{N} \right\} [/tex],

we can prove by use of the Archimedean property (given [tex]\delta[/tex] element of real such that [tex]\delta >0[/tex], there exist an n element of the positive intergers such that for any y element of real, [tex]n \delta > y[/tex]), that

[tex] \forall \delta>0, V'(1,\delta) \cap E \neq \emptyset [/tex] (the definition of 1 being an accumulation point)

Now I can't see how we can use the Archimedean property here. Anyone sees a way to do this?

Thanks for your inputs!