If $\limsup\limits_{n\rightarrow \infty} a_n=a< \infty$, then $\forall\epsilon>0$ $\exists N\in \mathbb{N}:a_n\leq a+\epsilon$

If $$\limsup\limits_{n\rightarrow \infty} a_n=a< \infty$$, then there exists for all $$\varepsilon >0$$ a $$N\in \mathbb{N}$$ such that $$a_n\leq a+\varepsilon$$ for all $$n\in \mathbb{N}$$, $$n\geq N$$

My attempt:

Suppose there are infinite elements $$a_{n_1},a_{n_2}, a_{n_3},...$$ with $$a_{n_k}\geq a$$. The sequence $$(a_{n_k})_k$$ is bounded above, otherwise $$(a_n)_n$$ wouldn't be bounded and $$\limsup\limits_{n\rightarrow \infty} a_n=+\infty$$. Hence $$(a_{n_k})_k$$ has - according to Weierstraß and Bolzano - a convergent subsequence $$(a_{n_{k_j}})$$ with a limit $$\geq a$$} since $$(a_{n_{k_j}})\geq a$$ for every $$j\in \mathbb{N}$$.

• How do you define $\limsup_na_n$? – José Carlos Santos Jun 2 at 12:03
• Like that: $\limsup\limits_{n\rightarrow \infty}a_n:=\begin{cases}\sup H, \text{ if } (a_n)_n \text{is bounded above} \\ \infty, \text{ else}\end{cases}$. $H$ is the set of limit points. – ParabolicAlcoholic Jun 2 at 12:06

Recall that $$\limsup\limits_{n\rightarrow \infty} a_n = \lim\limits_{n\to\infty} \sup\limits_{k\geq n} a_k$$. Let us fix $$\epsilon > 0$$. By definition of the limit, we know that there is some $$N\in \mathbb N$$ such that $$\sup\limits_{k\geq N} a_k < a+\epsilon$$. Thus, by definition of the supremum, we conclude that $$a_k < a + \epsilon$$ for every $$k\geq N$$, which is the conclusion you wanted.
Suppose otherwise. That is, suppose that, for each $$N\in\mathbb N$$, there is a $$n\in\mathbb N$$ such that $$n\geqslant N$$ and that $$a_n\geqslant a+\varepsilon$$. So, there is a sequence $$(a_{n_k})_{k\in\mathbb N}$$ such that $$(\forall k\in\mathbb N):a_{n_k}\geqslant a+\varepsilon$$. And, since $$(a_n)_{n\in\mathbb N}$$ is bounded, $$(a_{n_k})_{k\in\mathbb N}$$ is bounded too. So, it has a convergent subsequence, by the Bolzano-Weierstrass theorem. The limit of this subsequence must be greater than or equal to $$a+\varepsilon$$, but is is impossible, since, by definition, $$a$$ is the supremum of the set of limit points.