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Show that sequence is unbounded

by oceanthang
Tags: sequence, unbounded
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oceanthang
#1
Feb4-13, 04:48 AM
P: 3
1. The problem statement, all variables and given/known data
Show that the sequence an= sqrt(n) is unbounded


2. Relevant equations
there is no relevant equation require.


3. The attempt at a solution
Actually, I'm a newbie for real analysis, I try to prove it by using contradiction method,
but I stuck at half way, can anyone provide solution to me? thanks.
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Rikardus
#2
Feb4-13, 05:56 AM
P: 9
Suppose an is bounded. Then there exists M > 0 such that an <= M for all n. However this is absurd since you only have to pick n greater than M^2 to have an > M.

To illustrate: pick M = 120. 120^2 = 14400. a_14401 > 120.

Makes sense?
oceanthang
#3
Feb4-13, 08:21 PM
P: 3
Can you explain more detail about the illustration? I don't really get it about this statement
"only have to pick n greater than M^2 to have an > M", thanks.

Karnage1993
#4
Feb4-13, 08:31 PM
P: 132
Show that sequence is unbounded

Could you show us what you did to become stuck half way? We can assist you by checking if your approach was going the right way.
Rikardus
#5
Feb5-13, 05:09 AM
P: 9
Quote Quote by oceanthang View Post
Can you explain more detail about the illustration? I don't really get it about this statement
"only have to pick n greater than M^2 to have an > M", thanks.
Stating that a_n is bounded means that you can find a value of M which is greater than all the terms in the sequence. You must prove there is no such M. Reduction to the absurd works well in this situation.

You suppose there exists M >= a_n for all n. But there is a problem with this statement, which is a_{[M^2] + 1} > M, where [M^2] is the integer part of M squared.

Suppose you state that M = 120.2 is greater than all the terms in the sequence (a purely arbitrary choice).
That is absurd since a_{[120.2^2]+1} = a_14449 = sqrt(14449) > 120.2. This reasoning works for all real M.
oceanthang
#6
Feb5-13, 05:42 AM
P: 3
I finally get it, thank you~


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