- #1

cbarker1

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MHB

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- TL;DR 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.

Dear Everybody,

I have a quick question about the \\(\displaystyle M\\(\displaystyle in this proof:

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.

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

Thanks

C.barker\)\)\)\)\)\)\)\)\)\)\)\)\)\)

I have a quick question about the \\(\displaystyle M\\(\displaystyle in this proof:

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.

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

Thanks

C.barker\)\)\)\)\)\)\)\)\)\)\)\)\)\)