# lim sup

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• December 15th 2011, 01:04 AM
alphabeta89
lim sup
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?
• December 15th 2011, 03:36 AM
girdav
Re: lim sup
$\limsup_n x_n=\lim_n\sup_{k\geq n}x_k=\lim_n\max (\sup_{2k\geq n}x_{2k},\sup_{2k+1\geq n}x_{2k+1})$. Put the $\lim$ into the $\max$ to get the result.