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For example we have the two sequences

$A(n) = \dfrac{3n^2+n+2}{n^2-n+1}$

and

$ B(n) = 1+(-1/2) + (-1/2)^2 + (-1/2)^3 +\cdots+ (-1/2)^{n-1}$

I can't figure out the steps to work out the upper bound of these two sequences. I've searched so much but I couldn't find any clear examples of that type of questions about bounded sequences.

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So, I have edited this question and rendered equation in $\LaTeX$, Hope you'll implement them in your next posts –  user21436 Dec 27 '11 at 17:58
    
Thanks i didn't know how to do it –  Vaios Argiropoulos Dec 27 '11 at 17:59
    
If you have an upper bound on a sequence, then any larger number is also an upper bound for the sequence. Do you want the least upper bound? –  Dylan Moreland Dec 27 '11 at 18:18
    
Yeah i know that. I want some steps two prove both "general" bound and least upper bound. –  Vaios Argiropoulos Dec 27 '11 at 18:23
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3 Answers

up vote 1 down vote accepted

For $A(n)$, you probably know (or can believe) that $A(n)\to 3$ as $n\to\infty$. Hence $A(n)\le4$ for large $n$. More precisely, write $A(n) = 3+\dfrac{4n-1}{n^2-n+1}$ and find $N$ such that $4n-1\le n^2-n+1$ for all $n\ge N$. Then $A(n)\le 4$ for all $n\ge N$.

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For the first, it may help to look at $A(n+1) - A(n)$ and show that $A(n)$ is decreasing for $n \ge 2$.

For the second, do you know a formula for the sum of a geometric series?

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Generally, you can make a conjecture about the upper bound of the sequence (try looking at it like a regular function and taking the limit). For example, you can guess that $ A(n) \rightarrow 3 $ as $ n \rightarrow \infty $. Then you must show, by definition of the upper bound of a sequence, that

$ \forall\epsilon>0 $, $ \exists s(\epsilon)\in A(n): inf[A(n)]-s(\epsilon)<\epsilon$

,where $ inf[A(n)] $ is the upper bound. Does that make intuitive sense? If you take $ \epsilon $ to be given, it will help you to draw a picture on the number line to see exactly what you're looking for. As a further hint, it will help you to know that there is an Archimedean property that says $ \forall\epsilon\in\mathbb{R},\exists n\in\mathbb{N}:\frac{1}{n}<\epsilon $

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