# 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.

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@JulianKuelshammer Thank you, mind telling me why the special characters did not stick? Will promptly delete and rephrase. – js7354 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}\$.

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 thank you, much less complicated than I was making it out to be. – js7354 Oct 25 '12 at 22:52