1

Let $\{x_n\}$ be a bounded sequence of real numbers, and define a new sequence $\{\sigma_n\}$ by $$\sigma_n=\frac 1n\sum_{i=1}^nx_i.$$ Prove that $\limsup \sigma_n\le \limsup x_n$.

I am confused on how to attempt this problem. I don't see how the $\limsup x_n$ is greater then the arithmetic average of the $x_i$'s.

Martin Sleziak
  • 53,687
mocha
  • 26
  • Begin with: "Let $M=\limsup x_n$. Fix $\epsilon>0$. There is $N$ such that $x_i<M+\epsilon$ for all $n\ge N$...." Then relate the latter inequality to $\sigma_n$. –  Sep 25 '14 at 00:34
  • See http://math.stackexchange.com/questions/357693/inequality-concerning-limsup-and-liminf-of-cesaro-mean-of-a-sequence or http://math.stackexchange.com/questions/193157/if-sigma-n-fracs-1s-2-cdotss-nn-then-operatornamelim-sup-sigma – Martin Sleziak Sep 27 '14 at 06:09

0 Answers0