Find $\limsup$ and $\liminf$ of a sequence and prove $\liminf a_n \leq \limsup a_n$.

Question

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.

Answer 1

I suppose if you can see that $\liminf a_n = -1$ and $\limsup a_n = +1$, then trivially $\liminf a_n \leq \limsup a_n$.

Indeed for "any" sequence that is the case, the definition forces that, so it is among properties of limsup and liminf.

From Wikipedia http://en.wikipedia.org/wiki/Limit_superior_and_limit_inferior

Whenever $\liminf x_n$ and $\limsup x_n $both exist, we have $\liminf_{n \to \infty}x_n\leq \limsup_{n \to \infty}x_n$.

Answer 2

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, $L \leq S$.