Find accumulation points (real analysis)

[quasar987]

We must find the accumulation point for the set

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

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..

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

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

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

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

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

E = \left\{ \frac{1}{n} | n \in \mathbb{N} \right\}
or
E = \left\{ \frac{1}{2^n} | n \in \mathbb{N} \right\},

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

\forall \delta>0, V'(1,\delta) \cap E \neq \emptyset (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!

 

[Hurkyl]

Well, let's see. One thing you would like to do is to show that 3/(n^2+2) < \epsilon for some \epsilon that you have chosen, right?

Well, that's equivalent to showing that n^2 > (3 / \epsilon) - 2, right? Any ideas on using the archmedian property to show this?

 

[quasar987]

So you're saying that it would be okay, in order to prove that E has 1 as an accumulation point, to show that the terms,

\frac{3}{n + \frac{2}{n}}

and

\frac {3}{n^2 + 2}

have an accumulation point at 0. It crossed my mind but I didn't have any justification for it.


To answer your question: How about this way? We chose y to be 3 - 2\delta and since \delta>0, there exist n such that

\delta n>3 - 2\delta

divide both side by \delta and argue that if there exist such a n, then n^2 satisfy the inequality just has much

n^2 > \frac{3}{\delta} - 2

And now we can go back in time and find what we wanted, that is, \forall\delta>0,

\delta > \frac{3}{n^2 + 2} > 0

Awesome! Ok, for \frac {3}{n + \frac{2}{n}} now. Using your trick, we see that this is the same as showing that there exists n such that

n + \frac{2}{n} > \frac{3}{\delta}

and that this is the same as showing that there exist n such that

n > \frac{3}{\delta}

because if this is true, then for any, n, since n + \frac{2}{n}>n, our inequality is true too.

But proving that there exist n such that n > \frac{3}{\delta} is just basic Archemede with y = 3, so we're done there too.


Is this OK? Did you have something else in mind?

 

[Hurkyl]

Yep; it was this basic idea I wanted to motivate. This is the basic principle behind many arguments: looking at each contribution to the "error", and show they vanish.