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I'm taking my first calculus courses and I am just being introduced to such concepts as $\limsup$ and $\liminf$ , so from what I understood if we consider some convergent real sequence then all it's subsequences converge to the same limit, meaning $\limsup$ and $\liminf$ of this sequence do not exist? is this right? please correct me if I am wrong.

t_kln
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goatminam
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  • If a sequence converges then its liminf and limsup exist and are equal to its limit. Do you understand the definitions? – Karl Jan 10 '23 at 19:51
  • The limsup always exist in $[-\infty, +\infty]$. There is always a subsequence converging to the limsup. It is also the largest entity to which a subsequence can converge. Replace largest by smallest for the liminf. – Gribouillis Jan 10 '23 at 20:04

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If a sequence is convergent, then both {limes superior and limes inferior exists and are equal ( to the limit of the sequence ).

The interesting case is when a sequence has more than one accumulation points. In such cases limes superior is the biggest (if there is such) of the accumulation points and limes inferior is the smallest (if there is such). For example for $a_n = (-1)^n $ there are just two accumulation points, namely $-1,1$. Thus $\limsup a_n = 1$ and $\liminf a_n = -1$.

In fact exactly this property is what characterizes the convergent sequences ( in terms of limes superior and limes inferior) You can check out this , for example ,sequence converges iff limsup=liminf

Petar
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  • Oh, I see thank you. Another question if you don't mind, so in other complicated cases do we have then to find all the subsequential limits of a sequence to determine Lim Sup and Lim Inf ? – goatminam Jan 10 '23 at 19:58
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    In some very easy cases probably not a bad idea. In general I wouldn't say so. In fact it turns out the following: Given a bounded sequence, for example, ${a_n}$ the number $\lambda$ is called a nontrivial majorant, if the set of natural numbers ${n \in \mathbb{N} : a_n \leq \lambda '}$ is cofinite. In this case the set of non trivial majorants is bounded from below and its infimum is exactly the limes superior of ${a_n}.$ Also, something useful. If $c_n:=\sup{a_n,a_{n+1},\ldots}$ for all $n$, then $\lim\limits_{n \to \infty} c_n = \limsup\limits_{n \to \infty} a_n.$ – Petar Jan 10 '23 at 20:08