1. ## Sequences and Limits

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 $x>X$,

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

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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.

2. 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 $a_n = 0,3,0,3,0,3....$

3. Originally Posted by ThePerfectHacker
Let $a_n = 0,3,0,3,0,3....$

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.

4. 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?

5. 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.?
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.

6. 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
$\left| {a_j - 3} \right| = \left| {3 - 3} \right| = 0 < \varepsilon$.
$\left| {a_j - 3} \right| = \left| {0 - 3} \right| = 3 < \varepsilon$ - 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.