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

Question

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?

Answer 1

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 $\limsup b_n \le \lim a_n$.