Lim sups and lim infs homework

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Homework Help Overview

The discussion revolves around the concepts of limit superior (lim sup) and limit inferior (lim inf) in the context of sequences. Participants are examining the properties of a sequence \( a_n \) and its derived sequence \( b_m \), specifically focusing on the implications of boundedness and monotonicity.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are exploring the relationship between the convergence of the sequence \( a_n \) and the behavior of the sequence \( b_m \). There is uncertainty about how the definitions of \( b_m \) relate to its nonincreasing nature. Questions are raised regarding the assumptions made about the convergence of \( a_n \) and the implications for \( b_m \).

Discussion Status

Some participants have provided clarifications regarding the definitions and properties of lim sup, while others are still grappling with the implications of these definitions on the sequences involved. There is a recognition of differing interpretations of the definitions being used, particularly in relation to the specific problem parts.

Contextual Notes

There is a mention that the sequence \( a_n \) is not necessarily assumed to converge, which is relevant to the discussion of lim sup and lim inf. Participants are also considering the definitions of these concepts as they relate to the problem at hand.

koshn
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Homework Statement



http://i.imgur.com/ARy65.png

Homework Equations



Just general definitions of lim sups and lim infs

The Attempt at a Solution



I'm stuck on the part a. I realize that since a_n has a limit that means it is bounded and monotone. By I don't know how that implies that b_m is nonincreasing. The way I understand b_m is that b_1 = sup{a_m} and b_2 = sup{a_m,a_(m+1)} so is that correct? If so doesn't it depend on whether or not a_m is less than or greater than a_(m+1) and so on to know that b_m is nonincreasing?
 
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koshn said:
I'm stuck on the part a. I realize that since a_n has a limit that means it is bounded and monotone. By I don't know how that implies that b_m is nonincreasing. The way I understand b_m is that b_1 = sup{a_m} and b_2 = sup{a_m,a_(m+1)} so is that correct? If so doesn't it depend on whether or not a_m is less than or greater than a_(m+1) and so on to know that b_m is nonincreasing?

No, an is not assumed to converge (unless I missed something). The limsup is defined for any sequence, even divergent ones (although then sometimes the limsup may be infinity).

I don't think you totally understand the bm. The way bm is defined would give

[tex] b_1 = \sup\{a_1, a_2, a_3, a_4 \ldots\} [/tex]
[tex] b_2 = \sup\{a_2, a_3, a_4 \ldots\}[/tex]
[tex] b_3 = \sup\{a_3, a_4 \ldots\}[/tex]

,etc. Basically, as m increases, the first m terms of (an) are omitted when taking the sup. Now take another look at the hint. Does that help?
 
koshn - What definition of lim sup are you using? I ask because part (d) is often given as the definition of lim sup, whereas you are being asked to prove it.
 
spamiam said:
No, an is not assumed to converge (unless I missed something). The limsup is defined for any sequence, even divergent ones (although then sometimes the limsup may be infinity).

I don't think you totally understand the bm. The way bm is defined would give

[tex] b_1 = \sup\{a_1, a_2, a_3, a_4 \ldots\} [/tex]
[tex] b_2 = \sup\{a_2, a_3, a_4 \ldots\}[/tex]
[tex] b_3 = \sup\{a_3, a_4 \ldots\}[/tex]

,etc. Basically, as m increases, the first m terms of (an) are omitted when taking the sup. Now take another look at the hint. Does that help?

Thanks that helped to clear the confusion I was having.

jbunniii said:
koshn - What definition of lim sup are you using? I ask because part (d) is often given as the definition of lim sup, whereas you are being asked to prove it.

The definition I'm using is that lim sup x_n is defined to be the infimum of all numbers b with the following property: There is an integer N so that x_n < b for all n >= N.
 

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