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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...

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  • $\begingroup$ This is a special case of math.stackexchange.com/questions/122755/… $\endgroup$ – Listing Jan 8 '14 at 12:46
  • $\begingroup$ I'm confused...for each $n$ isn't $a_n$ a single number? so $\sup a_n = a_n$? $\endgroup$ – GPerez Jan 8 '14 at 12:53
  • $\begingroup$ @GPerez When taking the supremum, the domain is sometimes left unspecified when it is clear what it should be. In this case, $\sup a_{n}$ is the same as $\sup_{n\geq1}a_{n}$, as this is usually what one cares about when taking the supremum of a sequence. $\endgroup$ – Brian Jan 8 '14 at 13:01
  • $\begingroup$ Ohh that makes more sense, that means its the limit superior..I hadn't seen that notation before, I usually just write the whole definition out $\endgroup$ – GPerez Jan 8 '14 at 13:03
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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

$$\forall\;k>m\;\;:\;\;|a_{n_k}-L|<\epsilon \;\iff\;\;\lim_{k\to\infty}a_{n_k}=L\;\;\;\;\;\;\;\;\;\;\;\;\square$$

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  • $\begingroup$ So I think by showing that all subsequences have the same limit it follows that there is one limit point: the limit of the sequence, and therefore the limes superior is equal to that limit, right? $\endgroup$ – Spacer Jan 9 '14 at 12:06
  • $\begingroup$ Well, that is true but it is not what you asked. You asked to show that if the sequence converges then the lim sup of the sequence equals that limit, and the lemma above covers that since the lim sup is one of the possibly several partial limits of the sequence... $\endgroup$ – DonAntonio Jan 9 '14 at 12:40
  • $\begingroup$ Hmm I think it's a correct answer for the question if we use the definition as stated in the OP for the lim sup $\endgroup$ – Spacer Jan 9 '14 at 13:45

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