# limit superior ....

• May 28th 2010, 07:50 AM
flower3
limit superior ....
Find $\overline{lim} \ a_n$ where :
$a_n = \begin{cases}
1 &, \text{ if } n\ is \ square \\
0 &, \text{ if } n \ not \ square
\end{cases}$
• May 28th 2010, 02:57 PM
Drexel28
Quote:

Originally Posted by flower3
Find $\overline{lim} \ a_n$ where :
$a_n = \begin{cases}
1 &, \text{ if } n\ is \ square \\
0 &, \text{ if } n \ not \ square
\end{cases}$

Let $S$ represent the set of all subsequential limits of $a_n$. Evidently $S=\{0,1\}$ and so $\limsup\text{ }a_n=\sup\text{ }S=1$