# 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.

• It must be the other way around, $a_{odd}=\frac{1}{n}-1$ and $a_{even}=\frac{1}{n}+1$ – the8thone Dec 19 '13 at 2:26
• @Roozbeh-unity tahnks.. fixed!! – eChung00 Dec 19 '13 at 2:41
• See also: math.stackexchange.com/questions/353642/… (and other questions shown there among linked questions). – Martin Sleziak Jan 29 '15 at 20:12

I suppose if you can see that $\liminf a_n = -1$ and $\limsup a_n = +1$, then trivially $\liminf a_n \leq \limsup a_n$.
Whenever $\liminf x_n$ and $\limsup x_n$both exist, we have $\liminf_{n \to \infty}x_n\leq \limsup_{n \to \infty}x_n$.
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, $L \leq S$.