F flower3 Aug 2008 172 1 May 28, 2010 #1 Find \(\displaystyle \overline{lim} \ a_n \) where : \(\displaystyle a_n = \begin{cases} 1 &, \text{ if } n\ is \ square \\ 0 &, \text{ if } n \ not \ square \end{cases} \)

Find \(\displaystyle \overline{lim} \ a_n \) where : \(\displaystyle a_n = \begin{cases} 1 &, \text{ if } n\ is \ square \\ 0 &, \text{ if } n \ not \ square \end{cases} \)

Drexel28 MHF Hall of Honor Nov 2009 4,563 1,566 Berkeley, California May 28, 2010 #2 flower3 said: Find \(\displaystyle \overline{lim} \ a_n \) where : \(\displaystyle a_n = \begin{cases} 1 &, \text{ if } n\ is \ square \\ 0 &, \text{ if } n \ not \ square \end{cases} \) Click to expand... Let \(\displaystyle S\) represent the set of all subsequential limits of \(\displaystyle a_n\). Evidently \(\displaystyle S=\{0,1\}\) and so \(\displaystyle \limsup\text{ }a_n=\sup\text{ }S=1\)

flower3 said: Find \(\displaystyle \overline{lim} \ a_n \) where : \(\displaystyle a_n = \begin{cases} 1 &, \text{ if } n\ is \ square \\ 0 &, \text{ if } n \ not \ square \end{cases} \) Click to expand... Let \(\displaystyle S\) represent the set of all subsequential limits of \(\displaystyle a_n\). Evidently \(\displaystyle S=\{0,1\}\) and so \(\displaystyle \limsup\text{ }a_n=\sup\text{ }S=1\)