Sequences and Limits

**WWTL@WHL** · January 17th 2008, 10:05 AM

Question:
Give an example of a divergent sequence $(a_n)_{n\in\mathbb{N}}$ that satisfies the following conditions:

(i) For every $\epsilon >0$ , there exists an $X$ such that for infinitely many $x>X$ ,

$\left| a_n - 3 \right| < \epsilon$

(ii) There exists an $\epsilon >0$ and a $X \in\mathbb{N}$ such that for all $x>X$ ,

$\left| a_n - 3 \right| < \epsilon$

================================================== =======

I have absolutely no idea how to do these

For the infinitely many x>X one, I'm thinking this could involve a trig function since it doesn't converge and repeats every $2\pi$ . Please help.

**ThePerfectHacker** · January 17th 2008, 10:31 AM

Originally Posted by WWTL@WHL

Question:
Give an example of a divergent sequence $(a_n)_{n\in\mathbb{N}}$ that satisfies the following conditions:

(i) For every $\epsilon >0$ , there exists an $X$ such that for infinitely many $x>X$ ,

$\left| a_n - 3 \right| < \epsilon$

(ii) There exists an $\epsilon >0$ and a $X \in\mathbb{N}$ such that for all $x>X$ ,

$\left| a_n - 3 \right| < \epsilon$

Let $a_n = 0,3,0,3,0,3....$

**WWTL@WHL** · January 17th 2008, 11:31 AM

Originally Posted by ThePerfectHacker

Let $a_n = 0,3,0,3,0,3....$

Thanks for the reply.

But if I had a sequence like that: $a_n = 0$ for n is odd, and $a_n = 3$ for n is even....

...how does this satisfy the criteria of being less than $\epsilon$ ? I mean, if a_n = 0, then you have |-3| = 3 < $\epsilon$ and what if epsilon was 0.6, or 2 etc.?

Sorry if I'm being stupid.

**WWTL@WHL** · January 17th 2008, 12:25 PM

Hey, after spending the best part of an hour jotting stuff down on paper, I've come up with this for part (i)

$a_n = \frac{\epsilon}{2}\cos x + 3$

is that a divergent sequence? I think it is. And it satisfies the conditions, right? I mean if you let X = -1, then are infinitely many x>X.

no idea about (ii) though

Any ideas?

**Plato** · January 17th 2008, 12:39 PM

Originally Posted by WWTL@WHL

But if I had a sequence like that: $a_n = 0$ for n is odd, and $a_n = 3$ for n is even....
...how does this satisfy the criteria of being less than $\epsilon$ ? I mean, if a_n = 0, then you have |-3| = 3 < $\epsilon$ and what if epsilon was 0.6, or 2 etc.?

You seem to have misread your own question.
There are infinitely many even integers. So if j is even we have:
$\left| {a_j - 3} \right| = \left| {3 - 3} \right| = 0 < \varepsilon$ .

Moreover for the second condition, letting $\varepsilon = 4$ fulfills the condition.

**WWTL@WHL** · January 17th 2008, 12:50 PM

Originally Posted by Plato

Moreover for the second condition, letting $\varepsilon = 4$ fulfills the condition.

Indeed it does, thanks a lot mate!

Originally Posted by Plato

You seem to have misread your own question.
There are infinitely many even integers. So if j is even we have:
$\left| {a_j - 3} \right| = \left| {3 - 3} \right| = 0 < \varepsilon$ .

Yep, I did. Sorry about that.

But if j is odd we have:

$\left| {a_j - 3} \right| = \left| {0 - 3} \right| = 3 < \varepsilon$ - but what if epsilon was 2? That's the question I was trying to ask. There doesn't seem to be a way around it.

EDIT: Nevermind I understand it now. I've still been misreading my question. I think I get it now. Thanks Plato and ThePerfectHacker

Thread: Sequences and Limits

LinkBack

Thread Tools

Display

Sequences and Limits

Similar Math Help Forum Discussions

limits of sequences

Sequences and limits

Sequences - limits

Limits and sequences

limits of sequences

Search Tags