Proving a convergent sequence is bounded

[cbarker1]

TL;DR

Suppose \[math]b_n\[math] is in \[math]\mathbb{R}\[math] such that \[math]lim b_n=2\[math]. Proving the sequence is bounded.

Dear Everybody,

I have a quick question about the \[math]M\[math] in this proof:

Suppose \[math]b_n\[math] is in \[math]\mathbb{R}\[math] such that \[math]lim b_n=3\[math]. Then, there is an \[math] N\in \mathbb{N}\[math] such that for all \[math]n\geq\[math], we have \[math]|b_n-3|<1\[math]. Let M1=4 and note that for n\geq N, we have
|b_n|=|b_n-3+3|\leq |b_n-3|+|3|<1+3=M1. The set A= {|b_1|,|b_2|,...|b_{N-1}| is a finite set and hence let M2=max{A}. Then Let M=max{B1,B2}. Then for all n in N we have |b_n|\leq B.

Should M be a natural number or a real number? If real, why?

Thanks
C.barker[/math][/math][/math][/math][/math][/math][/math][/math][/math][/math][/math][/math][/math][/math]

 

[Office_Shredder]

It can be any real number.

 

[Mark44]

Summary: Suppose \[math]b_n\[math] is in \[math]\mathbb{R}\[math] such that \[math]lim b_n=2\[math]. Proving the sequence is bounded.

Suppose \(\displaystyle b_n\(\displaystyle is in \(\displaystyle \mathbb{R}\(\displaystyle such that \(\displaystyle lim b_n=3\(\displaystyle . Then, there is an \(\displaystyle N\in \mathbb{N}\(\displaystyle such that for all \(\displaystyle n\geq\(\displaystyle , we have \(\displaystyle |b_n-3|<1\(\displaystyle . Let M1=4 and note that for n\geq N, we have
|b_n|=|b_n-3+3|\leq |b_n-3|+|3|<1+3=M1. The set A= {|b_1|,|b_2|,...|b_{N-1}| is a finite set and hence let M2=max{A}. Then Let M=max{B1,B2}. Then for all n in N we have |b_n|\leq B.
[/math][/math][/math][/math][/math][/math]

@cbarker1, one of your LaTeX is displaying properly. Take a look at our LaTeX guide. The link is in the lower left corner of the text entry pane.

 

[WWGD]

Ultimately, you have finitely many terms $$a_n$$ that are not within a fixed $$\epsilon$$ of the limit $$L$$, and infinitely many within $$(L-\epsilon, L+ \epsilon)$$.