Let $(a_n)$ be a sequence of numbers. Show: If $(a_n)$ converges, than:
$\lim\limits \sup a_n= \lim\limits_{n \rightarrow \infty} a_n $
I can feel this is true intuitively, but I have no idea how to do formal proofs with lim sup as I've never really worked with the concept before. Can anyone give a helping hand?
Edit: Some more information:
The definition we've been thaught is that the lim sup is equal to sup V if the squence has an upperbound, with V being the set of all the limit points of the sequence. If the sequence has no upperbound lim sup is +infinity, and if the sequence has an upperbound but V is empty lim sup is -infinity.
So what I thought was if (An) converges than it must only have 1 limit point:lim (An). Therefor lim sup (An) must be equal to lim(An), because V only has 1 element. But I have no idea how to write this formally…
Best Answer
Hint:
Lemma: If a sequence converges (in the wide sense of the word) then any subsequence also converges and to the same limit.
Proof: In the finite limit case (the infinite limit case is left as an exercise):
$$\lim_{n\to\infty}a_n=L\implies \forall\,\epsilon>0\;\exists\,N_\epsilon\in\Bbb N\;\;s.t.\;\;n>N_\epsilon\implies|a_n-L|<\epsilon$$
Let now $\;\{a_{n_k}\}\subset\{a_n\}\;$ be a subsequence of our sequence. Remember that this means $\;n_1<n_2<\ldots\;$ . Thus, let $\;m\in\Bbb N\;$ be such that $\;n_m>N_\epsilon\;\implies\;n_k>N_\epsilon\;\;\forall\;k>m\;$ , so
$$\forall\;k>m\;\;:\;\;|a_{n_k}-L|<\epsilon \;\iff\;\;\lim_{k\to\infty}a_{n_k}=L\;\;\;\;\;\;\;\;\;\;\;\;\square$$