We must find the accumulation point for the set(adsbygoogle = window.adsbygoogle || []).push({});

[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!

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Find accumulation points (real analysis)

**Physics Forums | Science Articles, Homework Help, Discussion**