real analysis - bounded sequence and lim sup

Question

The question is

Let $(a_k)_k$ $_\in$ $_\mathbb N$ be a bounded sequence of real numbers.

Let b be a real number such that b > lim sup$_k$$_\to$$_\infty$ $a_k$

Prove that there exists N $\in$ $\mathbb N$ such that b > $a_k$ for all k > N.

Here is my attempt:

Denote $d_n$ = sup{$a_k$ | k $\geq$ n}

b > lim sup$_k$$_\to$$_\infty$ $a_k$ = $\lim_{n\to\infty}$ dn

$\exists$ N $\in \mathbb N$ such that $\forall n > N$, $d_n$ < b

$\exists$ $\epsilon>0$ such that b-$\epsilon$ > $\lim_{n\to\infty} dn$ = d [denote d = $\lim_{n\to\infty} dn$]

$\exists$ N $\in \mathbb N$ such that $\forall n \geq N$, |$d_n - d$| < $\epsilon$

$\Rightarrow$ $a_n \leq d_n < \epsilon + d = b$

Substitute n=k, we now have $\Rightarrow a_k < b$

Does this make sense?

Answer 1

Rephrasing:

$(a_k){k \in \mathbb{N}}$ is is bounded.

$d_n := \sup_{k \ge n} (a_k)$ is bounded and

decreasing, hence convergent.

Let $\lim_{n \rightarrow \infty} d_n =d$.

Given:

$b> \lim \sup_{n \rightarrow \infty} (a_n)=$

$\lim_{n \rightarrow \infty} d_n=d.$

Let $0< \epsilon \le b-d.$

There exists a $n_0 \in \mathbb{N}$ s.t. for

$n \ge n_0:$

$|d_n-d| <\epsilon$, or

$d_n \lt d +\epsilon \le d + b-d=b$.

Hence

$a_n \le d_n \lt b$.