Given that $(a_n)$ is a bounded sequence. If $b_n = \sup\{a_n, a_{n+1}, a_{n+2}, a_{n+3}, \ldots \}$ then does the limit always equal to $\sup\{a_n\}$ $\displaystyle\lim_{n\to\infty} b_n = \sup\{a_n\}$ My argument based on observation, $(b_n)$ is a decreasing sequence and bounded because $(a_n)$ is bounded. The limit exists and it's actually equal to the $\inf\{b_n\}$, but $(b_n)$ contains all the least upper bounds of subsequence of $(a_n)$, which implies it must be equal to $\sup\{a_n\}$.
However, the way I reason sounds a bit common sense, but I could not find any related theorem from my lecture notes so that I could give a more convincing argument. I wonder if anyone could give me some ideas on solving this problem? Thanks.
Question regarding the least upper bound of a bounded sequence.
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$\begingroup$
sequences-and-series
analysis
limits
1 Answers
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Suppose that $a_n=2^{-n}$ for $n\in\Bbb N$. Then $b_n=a_n$ for each $n\in\Bbb N$, so $\lim_{n\to\infty}b_n=0$, but $\sup_na_n=2^0=1$. Indeed, whenever $\langle a_n:n\in\Bbb N\rangle$ is strictly decreasing, you’ll have $\lim_{n\to\infty}b_n<\sup_na_n=a_0\;.$
$\lim\limits_{n\to\infty}b_n=\limsup_na_n$, which in general is not the same as $\sup_na_n$; for more information see here.
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0@Chan: You’re welcome! – 2012-10-13