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Thread: limit superior ....

  1. #1
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    limit superior ....

    Find $\displaystyle \overline{lim} \ a_n $ where :
    $\displaystyle a_n = \begin{cases}
    1 &, \text{ if } n\ is \ square \\
    0 &, \text{ if } n \ not \ square
    \end{cases} $
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by flower3 View Post
    Find $\displaystyle \overline{lim} \ a_n $ where :
    $\displaystyle a_n = \begin{cases}
    1 &, \text{ if } n\ is \ square \\
    0 &, \text{ if } n \ not \ square
    \end{cases} $
    Let $\displaystyle S$ represent the set of all subsequential limits of $\displaystyle a_n$. Evidently $\displaystyle S=\{0,1\}$ and so $\displaystyle \limsup\text{ }a_n=\sup\text{ }S=1$
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