cbarker1
Gold Member
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 \[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]
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]