# Thread: asking if this squeeze theorem proof is correct

1. ## asking if this squeeze theorem proof is correct

Hello,

I would like to ask if the proof below, concerning the squeeze theorem is correct. I guess someone else must have come up with it but I didn't find anything in inte rnet...

Gn =< An =< Bn ... where sequences (Gn), (Bn) -> a, then a is a limit for (An) as well.
[=< less or equal, -> converges to, <=> equivalent]

Proof
-Consider sequences (Xn), (Yn) of positive or zero terms! (1)

Gn=<An <=> Gn=An-Xn <=> |Gn-a|=|An-Xn-a| <=> |An-Xn-a|<e <=> |An-(a+Xn)|<e <=> An->(a+Xn)

Bn>=An <=> Bn=An+Yn <=> |Bn-a|=|An+Yn-a| <=> |An+Yn-a|<e <=>
|An -(a-Yn)|<e <=> An->(a-Yn)

This is:
{An->(a+Xn) , An->(a-Yn)} => a+Xn=a-Yn <=> Xn=Yn=0 because follows from (1) they are positive or zero.

Please, if anyone finds that there is a mistake, or it is correct, I would be very glad to know.

Thank you all, in advance!

2. Originally Posted by Melsi
Hello,

I would like to ask if the proof below, concerning the squeeze theorem is correct. I guess someone else must have come up with it but I didn't find anything in inte rnet...

Gn =< An =< Bn ... where sequences (Gn), (Bn) -> a, then a is a limit for (An) as well.
[=< less or equal, -> converges to, <=> equivalent]

Proof
-Consider sequences (Xn), (Yn) of positive or zero terms! (1)

Gn=<An <=> Gn=An-Xn <=> |Gn-a|=|An-Xn-a| <=> |An-Xn-a|<e <=> |An-(a+Xn)|<e <=> An->(a+Xn)

This makes no sense: in order to obtain the limit of $A_n$ you must make $n\to\infty$ , so
how

come n still appears in the expression $A_n\xrightarrow [n\to\infty]{} a+X_n$ ??!

The same can be said about the following step, too.

Tonio

Bn>=An <=> Bn=An+Yn <=> |Bn-a|=|An+Yn-a| <=> |An+Yn-a|<e <=>
|An -(a-Yn)|<e <=> An->(a-Yn)

This is:
{An->(a+Xn) , An->(a-Yn)} => a+Xn=a-Yn <=> Xn=Yn=0 because follows from (1) they are positive or zero.

Please, if anyone finds that there is a mistake, or it is correct, I would be very glad to know.

Thank you all, in advance!
.

3. Thank you very much for your answer. It is great to have your thought checked, I appreciate it.

Having in mind not to waste yours or others time I would like to make a final try and chang it a bit:

Gn=< An <=> Gn=An-Xn <=>|(An-Xn)-a|<e after suitable index <=>(An-Xn)->a

Bn>= An <=> Bn=An+Yn <=>|(An+Yn)-a|<e after suitable index <=>(An+Yn)->a

1. are the above two lines correct enough in order to make the consumption that Xn=Yn=0 then An->a?
2. isn't the random positive e enough to make the index n to run infinitely.

Thank you very much for your help!

4. I think that your approach will not lead you to a proof.
Here is an outline which is very short on specifics.

Suppose that $\epsilon>0$.
By carefully selecting indices we can get the following.
$|B_n-G_n|\le |B_n-a|+|a-G_n|<\frac{\epsilon }{4}+\frac{\epsilon }{4} =\frac{\epsilon }{2}$.

Now $0<|A_n-G_n|\le |B_n-G_n|<\frac{\epsilon }{2}$

So $|A_n-a|\le |A_n-G_n|+|G_n-a|<\epsilon$.

5. ## Very good solution!

Well, I think this one is excellent, I have already analyzed it in my notebook and it is very satisfying.

I am very pleased with the answer and so grateful,
Thank you very much!