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- Thread starter mynameisfunk
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- #1

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- #2

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Edit: Misread the question.

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- #3

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wouldnt that make (a) false then? but a goes to infinity.... truncated on infinity?

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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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[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)`

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with the inequality [tex]m \leq n [/tex]

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Code:

`\mathop {\lim }\limits_{n \to \infty}} \mathop {\sup }\limits_{m \geq n}} s_m`

- #9

HallsofIvy

Science Advisor

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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

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

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Thanks

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