# Supremum of a series

Let $( x_n )$ be a bounded sequence, and define for each $m \in \Bbb{N}$, the number $u_m := \sup{x_n : n \le m}$. Show that $(u_m)$ is a bounded, decreasing sequence. I'm having trouble even conceptualizing this question, specifically, how to define $(x_n)$ in terms of each ($m \in \Bbb N$). Any insight would be much appreciated.

-
Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. If this is homework, please add the homework tag; people will still help, so don't worry. Also, many find the use of imperative ("Prove", "Solve", etc.) to be rude when asking for help; please consider rewriting your post. – Julian Kuelshammer Oct 25 '12 at 22:36
@JulianKuelshammer Thank you, mind telling me why the special characters did not stick? Will promptly delete and rephrase. – kqualters Oct 25 '12 at 22:37
You need to add $\$$-signs to indicate that you are in math mode. – Julian Kuelshammer Oct 25 '12 at 22:39 For more basic information about writing math at this site see e.g. here, here, here and here. – Julian Kuelshammer Oct 25 '12 at 22:39 ## 1 Answer As (x_n) is bounded, there exist C < \infty such that |x_n|<C. Then u_m<C. Notice that$$ \{ x_n : n\le m\} \subset \{x_n : n \le m+1\}.$$Then,$u_m\le u_{m+1}\$.

-
thank you, much less complicated than I was making it out to be. – kqualters Oct 25 '12 at 22:52