Find $\limsup$ and $\liminf$ of a sequence and prove $\liminf a_n \leq \limsup a_n$. I have a question of finding lim sup and lim inf of $a_n=\frac{1}{n} + (-1)^n$ and prove $\liminf  a_n \leq \limsup  a_n.$
So the work below is what I did for the first part.
$a_{odd\ n} = \frac{1}{n}-1$ and $a_{even\ n} = \frac{1}{n}+1.$ So $\limsup  a_n = 1$ and $\liminf  a_n = -1.$
How do I prove the second part?? I tried to use the definition but I am confused with the definition.

Here is how to prove $\liminf \{ x_n \} \leq \limsup \{ x_n \}$ for a general real sequence $\{ x_n \}$.  Set $L = \liminf \{ x_n \}$ and $S = \limsup \{ x_n \}$, and suppose for the sake of contradiction that $L > S$.  Say $L = S + h$ for some $h > 0$.  Then there are infinitely many $n$ for which $x_n \in (L - h, L + h)$.  Therefore, there are infinitely many $n$ for which $x_n > S$, contradicting the definition of $S$.  Hence, $L \leq S$.
