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.
This is a question from the book Methods of Real Analysis by R. R. Goldberg.     If $(s_n)$ is a sequence of real numbers and if $$\sigma_n=\frac{s_1+s_2+\cdots+s_n}{n}$$ then prove that: $\operatorname{{lim sup}}\sigma_n \leq \operatorname{lim sup} s_n$.   I don't have any idea how to start working on this problem. Please help. Thanks.
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