# Math Help - Sequence Proof (Hard)

1. ## Sequence Proof (Hard)

Hi! I need some help with this question:

(a)
Statement reads: For every $\epsilon$ there exists a point in the sequence $N \in N$ after which all terms lie between the lines $y= L-\epsilon$ and $y= L+ \epsilon$.

Now if we interchange the quantifiers it reads: There exists $N \in N$for every $\epsilon > 0$ such that for every point greater than or equal to N, all terms are between the lines $y= L-\epsilon$ and $y= L+ \epsilon$.

(sigh), it's the same thing! What's the difference if we interchange the two quantifiers? And how do we go about proving it? I tried everything, even drew a diagram. But man it’s hard...

So any help here is appreciated.

2. Hello Roam
Originally Posted by Roam
Hi! I need some help with this question:

(a)
Statement reads: For every $\epsilon$ there exists a point in the sequence $N \in N$ after which all terms lie between the lines $y= L-\epsilon$ and $y= L+ \epsilon$.

Now if we interchange the quantifiers it reads: There exists $N \in N$for every $\epsilon > 0$ such that for every point greater than or equal to N, all terms are between the lines $y= L-\epsilon$ and $y= L+ \epsilon$.

(sigh), it's the same thing! What's the difference if we interchange the two quantifiers? And how do we go about proving it? I tried everything, even drew a diagram. But man it’s hard...

So any help here is appreciated.
No, it isn't the same thing at all. In the first statement, we choose $\epsilon$ first, and then find the $N$ for which $|a_n -L|<\epsilon$ whenever $n\ge N$. In other words, the value of $N$ depends upon the value of $\epsilon$.

In the second statement, there is a fixed value of $N$, such that whenever $n\ge N, |a_n-L|$ can be made arbitrarily small.

3. Thanks Grandad, I see the difference now.

But I still don't know what method to use in order to prove the second statement. Could you give me some clues?

4. Originally Posted by Roam
Thanks Grandad, I see the difference now.

But I still don't know what method to use in order to prove the second statement. Could you give me some clues?
Notice: $|a_{n}|=|(a_{n}-a)+a|\leq\\|a_{n}-a|+|a|$. If we subtract $|a|$ form both sides, we have that, $|a_{n}|-|a|\leq\\|a_{n}-a|$. Thus, $|a_{n}-a|\geq\\||a_{n}|-|a||$.

Now for the proof...

Pick and arbitrary $\epsilon>0$. Since $(a_{n})$ converges to some limit a, there exists a natrual number $N$ such that, whenever $n\geq\\N$ implies $|a_{n}-a|<\epsilon$. However, $||a_{n}|-|a||\leq\\|a_{n}-a|<\epsilon$ also. Therefore, $||a_{n}|-|a||<\epsilon$.

5. Originally Posted by Danneedshelp
Notice: $|a_{n}|=|(a_{n}-a)+a|\leq\\|a_{n}-a|+|a|$. If we subtract $|a|$ form both sides, we have that, $|a_{n}|-|a|\leq\\|a_{n}-a|$. Thus, $\color{red}|a_{n}-a|\geq\\||a_{n}|-|a||$.
You have skipped a vital step there
See this proof.

6. Originally Posted by Plato
You have skipped a vital step there
See this proof.
Thanks,

I just skipped to the result needed for the proof, but thanks for pointing that out.

7. Originally Posted by Danneedshelp
Notice: $|a_{n}|=|(a_{n}-a)+a|\leq\\|a_{n}-a|+|a|$. If we subtract $|a|$ form both sides, we have that, $|a_{n}|-|a|\leq\\|a_{n}-a|$. Thus, $|a_{n}-a|\geq\\||a_{n}|-|a||$.

Now for the proof...

Pick and arbitrary $\epsilon>0$. Since $(a_{n})$ converges to some limit a, there exists a natrual number $N$ such that, whenever $n\geq\\N$ implies $|a_{n}-a|<\epsilon$. However, $||a_{n}|-|a||\leq\\|a_{n}-a|<\epsilon$ also. Therefore, $||a_{n}|-|a||<\epsilon$.
Hi!

I was thinking of using a proof by contradiction. I'm not sure how to do it but something like:

Suppose $a_n \neq L$ for any $n \geq N$. And then somehow try to prove that this assumption violates the condition:

$
\exists N \in \mathbb{N} \forall \epsilon > 0 \forall n \geq N |a_n -L|< \epsilon
$

Is this a good strategy?