I'm asking this to see if my reasoning makes sense. Assume $\{a_n\}$ is a bounded sequence, then $\lim\sup$ and $\lim\inf$ exist. I'll show this for $\lim\sup$ as it seems pretty straight forward:

Because there exists a least upper bound $a$, then $a_i \leq a$ for all $i$. Thus for all $\epsilon>0$ we see that $a_i< a + \epsilon$. Assume that there exists some $b$ such that $a_i < b + \epsilon < a + \epsilon$. Then $b < a$. This is a contradiction as $a$ is the least upper bound. Thus $a$ is the smallest real number such that for all $\epsilon >0$, $a_i<a+\epsilon$ for all $i$. Thus for all $\epsilon >0 $ there exists $N\in\mathbb{N}$ such that $a_n<a+\epsilon$ for all $n>N$. Thus $a+\epsilon$ is the $\lim\sup$.



1 Answer 1


If we denote $A$ the set of sequential limits of a bounded sequence $a_n$ then it is proved that $\limsup a_n=\max A$.

$(1)$If $b_n \to b$ and $b_n \leq a,\forall n \in \mathbb{N}$ for some $a \in \mathbb{R}$ then $\lim_{n \to +\infty} b_n \leq a$

Now $a_n \in [-M,M]$ where $M=\sup_{n \in \mathbb{N}}$ because it is assumed bounded.

Now exists $a_{k_n} \to \limsup a_n$ because limsup is a sequential limit.

Now $a_{n_k} \in [-M,M]$ thus from $(1)$ we have that $\limsup a_n \in [-M,M]$

so $\limsup a_n=a \in [-M,M]$.

Also from this you see that limsup cannot be larger than $\sup a_n +\epsilon$ for any $\epsilon>0$

  • $\begingroup$ Thanks! I realized after a bit that I was thinking of $\lim \sup$ wrong. Your answer is still greatly appreciated $\endgroup$
    – sadlyfe
    Sep 19, 2017 at 18:06
  • $\begingroup$ glad i could help $\endgroup$ Sep 19, 2017 at 18:21

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