Find limsup and liminf of a sequence From Probability through problems by Marek Capinski,Jerzy Zastawniak,:
Find $\limsup_{n \to \infty}A_n$ and $\liminf_{n \to \infty}A_n$,where 
\begin{eqnarray*}
A_n &=&\left(\frac 13-\frac1{n+1},1+\frac 1{n+1}\right) \mbox{ if}\space n=1,3,5,... \\&=&\left(\frac 1{n+1},\frac 23-\frac{1}{n+1}\right) \space\mbox{ if} \space n=2,4,6,...\\ \end{eqnarray*}
We are given definition as:
$\limsup_{n \to \infty}A_n=\cap_{n=1}^{\infty}\cup_{k=n}^{\infty}A_k$
$\liminf_{n \to \infty}A_n=\cup_{n=1}^{\infty}\cap_{k=n}^{\infty}A_k$
I am unable to use this definition to find the $\limsup_{n \to \infty}A_n$ and $\liminf_{n \to \infty}A_n$ of above sequence.Please let me know how can I get the required limits.
Thanks in advance!
 A: A way to characterize $\limsup A_n$ is:$$x\notin\limsup A_n\iff\{n\mid x\in A_n\}\text{ is finite}\tag1$$
A way to characterize $\liminf A_n$ is:$$x\in\liminf A_n\iff\{n\mid x\notin A_n\}\text{ is finite}\tag2$$
Observing that $$\{n\mid x\notin A_n\}\text{ is finite}\implies \{n\mid x\in A_n\}\text{ is infinite}$$ we conclude that: $$\liminf A_n\subseteq\limsup A_n$$
With tools $(1)$ and $(2)$ check out for some $x\in\mathbb R$.  This by discerning the following cases:


*

*$x\leq0$

*$0<x<\frac13$

*$\frac13\leq x<\frac 23$

*$\frac23\leq x<1$

*$x\geq1$
A: There is another definition that might help you in this case : limsup and liminf can be interpreted as (respectively) the sup and inf of the set of adherence values. More on that on Wikipedia.
Here, your adherence values would be $(1/3,1)$ and $(0,2/3)$. Does that help?
A: We can start by the $\liminf$. Let $x\in\mathbb R$. Saying that $x\in\liminf A_n$ means that there exists $N$ such that $x\in A_n$ for all $n\geqslant N$. In particular, for $n$ large enough, $x\in A_{2n}$ and $x\in A_{2n+1}$. This information give that a bound for $x$ depending on $n$. Letting $n$ going to infinity gives a bound independent on $n$ (you should get actually an interval). Then it remains to check that each element of this interval belongs to $\liminf A_n$.
For the $\limsup$, it can help to have a nice characterization of $\bigcup_{k=n}^{+\infty}A_k$ for a fixed $n$. To do so, split the union according to the indices where $k$ is odd or even.
