Confused by this question (adv calc)

  • #1
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Homework Statement



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I am confused by what they mean on part (a) are these sequences [tex]s_m[/tex] and [tex]s_n[/tex] the same sequence and this is what i should be showing?
 

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Answers and Replies

  • #2
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Edit: Misread the question.
 
Last edited:
  • #3
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wouldnt that make (a) false then? but a goes to infinity.... truncated on infinity?
 
  • #4
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Misread the question. Thought it said m < n... Sorry.
 
  • #5
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also, if someone could show me show to write part a in LateX i would appreciate it
 
  • #6
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I don't know the answer to the question but I can show you the latex.

[tex] \mathop {\lim \sup }\limits_{n \to \infty}} s_n = \mathop {\lim}\limits_{n \to \infty}} \left( \sup \lbrace s_m | m \geq n \rbrace \right) [/tex]

Code:
 \mathop {\lim \sup }\limits_{n \to \infty}} s_n = \mathop {\lim}\limits_{n \to \infty}} \left( \sup \lbrace s_m | m \geq n \rbrace  \right)
 
  • #7
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with the inequality [tex]m \leq n [/tex]
 
  • #8
128
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[tex]\mathop {\lim }\limits_{n \to \infty}} \mathop {\sup }\limits_{m \geq n}} s_m[/tex]

Code:
\mathop {\lim  }\limits_{n \to \infty}} \mathop {\sup }\limits_{m \geq n}} s_m
 
  • #9
HallsofIvy
Science Advisor
Homework Helper
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Yes, [itex]s_n[/itex] and [itex]s_m[/itex] refer to the same sequence, just numbered differently.

As for [itex]\lim_{n\to\infty} sup s_n= \lim_{n\to\infty}\{sup s_m|m\ge n\}[/itex]
look at some simple examples. if n= 1, then [itex]\{s_m|m\ge 1\}[/itex] is just the sequence itself. If n= 2, then [itex]\{s_m|m\ge 2\}[/itex] is all terms of the sequence except the first. In general, [itex]\{s_m|m\ge n\}[/itex] is the set of all terms of the sequence except those before [itex]s_n[/itex]. "[itex]\{sup s_n|m\ge 1\}[/itex] is the supremum (greatest lower bound) of all terms in the sequence beyond a certain point.

In a special case, suppose [itex]\{a_n\}[/itex] converges to A. Then, given [itex]\epsilon> 0[/itex] there exist N such that if n> N, [itex]|a_n- A|< \epsilon[/itex] so there are no members of the sequence, beyond n= N, that are larger than [itex]A+ \epsilon[/itex] and it is easy to see that, as n goes to infinity, the supremums of [itex]\{s_n|m\ge n\}[/itex] must go to A.

Suppose the sequence has a subsequence that converges to A and a subsequence that converges to B. Then no matter how large n is, there exist numbers in the sequence beyond n that are close to A and numbers close to B. The supremum will be close to whichever of A or B is larger. In the limit, the supremum will be the larger of A and B.

In general lim sup is the supremum of the set of all sub-sequential limits of the sequence. That is, you determine all numbers to which sub-sequences converge and find their supremum.
 
  • #10
125
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Thanks
 

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