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sequence a(n) is bounbed and sequence b(n) converges. Show that
a(n)≤b(n) (∀ n ∈ ℕ) ⇒lim sup a(n)≤lim b(n)

Since a(n) is bounded, a(n) has a convergent subsequence. let a1'(n) be a subsequence of a(n). Then, a1'(n)≤b(n) (∀ n ∈ ℕ), a2'(n)≤b(n) (∀ n ∈ ℕ),... ak'(n)≤b(n) (∀k∈ ℕ) thus, lim ak'(n)≤lim b(n) ⇒lim sup a(n)≤lim b(n) Is this a valid proof?

Martin Sleziak
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Roland
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    For some basic information about writing math at this site see e.g. here, here, here and here. – Jesse P Francis Jun 08 '15 at 10:44
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    How do you define $\limsup$? – Berci Jun 08 '15 at 10:44
  • See http://math.stackexchange.com/questions/213719/prove-a-n-leq-b-n-implies-limsup-a-n-leq-limsup-b-n and other quesions linked there. – Martin Sleziak Jun 08 '15 at 18:08
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1 Answers1

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I don't think that it is a valid proof.

First of all, how do you define $a_2^\prime(n)$?

Second, it is not because a subsequence of a sequence is always less that another sequence that $\limsup$ is also less.

A valid proof would be:

  • As for all $n \in \mathbb{N}$ $b_n \le a_n$, you have $\limsup b_n \le \limsup a_n$. Do you know this result? Can you prove it?
  • As $(a_n)$ is convergent, $\limsup a_n = \lim a_n$.
  • Therefore $\limsup b_n \le \lim a_n$.
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