I know it is a basic question but the definitions in POMA are too rigorous.

I need some sort of example to understand what's going on.

Can someone please help?

Thank you so much

  • 4
    $\begingroup$ Think of $(-1)^n$ $\endgroup$ – Prahlad Vaidyanathan Dec 5 '13 at 12:15
  • 2
    $\begingroup$ If the limit exists then it equals the limsup and liminf. So to notice the difference you must look at cases where the limit does not exist. See comment of @PrahladVaidyanathan. $\endgroup$ – drhab Dec 5 '13 at 12:17

Given a sequence $(a_n )_n$, a number $\ell$ is called a partial limit of $(a_n)_n$ if there exists a subsequence $\left(a_{n_k}\right)_k$ which converges to $\ell$. It can be shown that the set $$S=\{ \ell: \ell \text{ is a partial limit of }(a_n) \}$$ is closed, which furnishes the existence of $\max S$ and $\min S$. These are known as the limes superior and the limes inferior of $(a_n)$ respectively, or $\limsup a_n,\liminf a_n$. Note that the two always exist!

Now, in the special case that $\liminf a_n=\limsup a_n$ we have that all partial limits of $(a_n)$ (elements of $S$ defined above) coincide to a single number. In this case we say that $(a_n)$ itself has a limit, which is defined to be this common value.

EXAMPLE: The sequence $(a_n)=(-1)^n$ has partial limits $\pm 1$ obtained by the subsequences of odd/even indices. We thus have $$S=\{-1,+1\} $$ and $$\liminf a_n=-1 \\ \limsup a_n=+1. $$ In this case the sequence $(a_n)$ itself has no limit, also called divergent.

  • $\begingroup$ In the case of $a_n = n \sin(n)$ what is the set $S$ ? It is empty or $\lbrace -\infty, +\infty\rbrace$. Maybe you should remark that the maximum and minimum of $S$ are defined provided that the sequence is bounded above and below respectively. $\endgroup$ – Bunder Dec 5 '13 at 13:13
  • $\begingroup$ @Bunder Instead, I suggest you take the closure in the extended real line $\bar{\mathbb R}=\mathbb R \cup \{ \pm \infty \}$. $\endgroup$ – user1337 Dec 5 '13 at 13:30

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