# Proving a convergent sequence is bounded

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• cbarker1

#### cbarker1

Gold Member
MHB
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$$\)\)\)\)\)\)\)\)\)\)\)\)\)

Summary: Suppose \$$\displaystyle b_n\\(\displaystyle is in \\(\displaystyle \mathbb{R}\\(\displaystyle such that \\(\displaystyle lim b_n=2\\(\displaystyle . 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.$$\)\)\)\)\)
Ultimately, you have finitely many terms $$a_n$$ that are not within a fixed $$\epsilon$$ of the limit $$L$$, and infinitely many within $$(L-\epsilon, L+ \epsilon)$$.