Proving a convergent sequence is bounded

In summary, the conversation discusses the proof of a sequence being bounded, with the use of limits and finite and infinite sets. The question posed is whether the bound M should be a natural or real number.
  • #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\)\)\)\)\)\)\)\)\)\)\)\)\)\)
 
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  • #2
It can be any real number.
 
  • #3
cbarker1 said:
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\(\displaystyle . Let M1=4 and note that for n\geq N, we have
|b_n|=|b_n-3+3|\leq |b_n-3|+|3|<1+3=M1. The set A= {|b_1|,|b_2|,...|b_{N-1}| is a finite set and hence let M2=max{A}. Then Let M=max{B1,B2}. Then for all n in N we have |b_n|\leq B.
\)\)\)\)\)\)
@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.
 
Last edited:
  • #4
Ultimately, you have finitely many terms $$a_n$$ that are not within a fixed $$\epsilon$$ of the limit $$L$$, and infinitely many within $$(L-\epsilon, L+ \epsilon)$$.
 

What is a convergent sequence?

A convergent sequence is a sequence of numbers that approaches a specific limit as the sequence continues. In other words, as you go further down the sequence, the numbers get closer and closer to a certain value.

Why is it important to prove that a convergent sequence is bounded?

Proving that a convergent sequence is bounded is important because it guarantees that the sequence has a finite limit. This allows us to make accurate predictions and calculations based on the sequence.

What does it mean for a sequence to be bounded?

A bounded sequence is one in which all of the terms in the sequence fall within a certain range of values. This means that the sequence does not have any terms that are infinitely large or infinitely small.

How do you prove that a convergent sequence is bounded?

To prove that a convergent sequence is bounded, you must show that there exists a number M such that all terms in the sequence are less than or equal to M. This can be done using the definition of convergence and the properties of real numbers.

What are some common techniques for proving that a convergent sequence is bounded?

Some common techniques for proving that a convergent sequence is bounded include using the definition of convergence, applying the triangle inequality, and using the squeeze theorem. It is also helpful to use known properties of real numbers, such as the fact that the sum of two bounded numbers is also bounded.

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