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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 $j\in \mathbb{N}$.

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  • $\begingroup$ How do you define $\limsup_na_n$? $\endgroup$ – José Carlos Santos Jun 2 at 12:03
  • $\begingroup$ 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. $\endgroup$ – ParabolicAlcoholic Jun 2 at 12:06
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You don't need to go as far as using Bolzano-Weierstraß theorem. We can prove your statement directly.

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.

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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.

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