Find accumulation points (real analysis)

Homework Helper
Gold Member
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?

Hurkyl
Staff Emeritus
Gold Member
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?

Homework Helper
Gold Member
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
Staff Emeritus