# real analysis - bounded sequence and lim sup

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?

• To much abusive quantification ! As you said, there is $N$ s.t. $d_n<b$ for all $n\geq N$. In particular, $a_k\leq d_n$ for all $k\geq N$. The claim follow.
– Surb
Oct 17, 2018 at 9:02
• I would add more words and maybe put subscripts on each of the N variables, but the idea is correct. Oct 17, 2018 at 9:07

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

$$a_n \le d_n \lt b$$.

• does the epsilon has to be less than b-d? other than that, the rest seems more clear than my attempt
– TUC
Oct 17, 2018 at 9:48
• Tuc.$\epsilon = b-d$ is fine, this still gives in second last line $d_n <b$.Is this what you meant? Oct 17, 2018 at 9:54
• yes! thanks a lot
– TUC
Oct 17, 2018 at 9:58
• Tuc.A pleasure.I fix it.:) Oct 17, 2018 at 10:01
• Sorry but just 1 last question. Epsilon is less than or equal to b-d I stood that. But epsilon still has to be bigger than 0 right? I'm quite new to real analysis and take some time to understand... sorry!
– TUC
Oct 17, 2018 at 10:18