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Thread: Prove that lim sup(x_n) = max(lim sup(y_n), lim sup(z_n))

  1. #1
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    Prove that lim sup(x_n) = max(lim sup(y_n), lim sup(z_n))

    Hi there, my first question here. Hope someone can give me some hints on it. Thanks!

    Let $\displaystyle (x_{n})$ be a bounded sequence. For each $\displaystyle n \in \mathbb{N}$, let $\displaystyle y_{n}=x_{2n}$ and $\displaystyle z_{n}=x_{2n-1}$. Prove that
    $\displaystyle \lim \sup {x_n} = \max (\lim \sup {y_n},\lim \sup {z_n})$
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  2. #2
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    Re: Prove that lim sup(x_n) = max(lim sup(y_n), lim sup(z_n))

    To show that $\displaystyle \limsup x_n = L$ (I'm using L to represent $\displaystyle \max(\limsup y_n,\limsup z_n)$), you need to show that $\displaystyle x_n < L$ for all n, and for every $\displaystyle \epsilon > 0$, there is an n such that $\displaystyle x_n > L-\epsilon$. Can you see how to show each part?

    - Hollywood
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