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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.

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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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  • $\begingroup$ Thanks a lot Brian. I see it now, I really messed up on this problem badly. I can prove that $\lim_{n\to\infty}b_n$ converge, but I'm stuck with finding what does it converges to. $\endgroup$ – Chan Oct 13 '12 at 5:26
  • $\begingroup$ @Chan: $\lim_{n\to\infty}b_n$ depends very much on the specific sequence $\langle a_n:n\in\Bbb N\rangle$ that you start with. What you can prove is that if $\lim_{n\to\infty}b_n=b$, then for every $\epsilon>0$ there are only finitely many $n$ such that $a_n>b+\epsilon$. In other words, every real number $>b$ is an upper bound for some tail of the sequence $\langle a_n:n\in\Bbb N\rangle$. $\endgroup$ – Brian M. Scott Oct 13 '12 at 5:31
  • $\begingroup$ @Chan: You’re welcome! $\endgroup$ – Brian M. Scott Oct 13 '12 at 7:38

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