limit superior of sequence

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• January 12th 2012, 08:10 PM
alphabeta89
limit superior of sequence
Let $(x_n)$ be a bounded sequence. For each $n \in \mathbb{N}$ , let $y_n=x_{2n}$ and $z_n=x_{2n-1}$. Prove that
$\limsup{x_n}=\max({\limsup{y_n},\limsup{z_n}})$.

I tried using the identity $\max(x,y)=\frac{1}{2}{(x+y-|x-y|)}$, but it can't seem to work... can anyone help me out?
• January 13th 2012, 02:24 PM
girdav
Re: limit superior of sequence
Already discussed here.