Let $(s_n)$ be a sequence of nonnegative numbers, and $\sigma_n=\frac{1}{n}(s_1+s_2+\cdots +s_n)$. Show that $\liminf s_n \le \liminf \sigma_n$. Let $(s_n)$ be a sequence of nonnegative numbers, and for each $n$ define $\sigma_n=\frac{1}{n}(s_1+s_2+\cdots +s_n)$. 
Show that $\liminf s_n \le \liminf \sigma_n$.
Actually, I showed $\limsup \sigma_n \le \limsup s_n$ in the following way. 
For any $n \gt M \gt N$, we can get the following inequality, 
$\sup \{\sigma_n: n\gt M\}\le \frac{1}{M}(s_1+s_2+\cdots +s_N)+\sup\{s_n:n\gt N\}.$
So first taking the limit as $M\to \infty$ then as $N \to \infty$, we get the inequality. 
However, this method does not work out for the $\liminf$ case, since the above inequality was derived using the fact that $1/n \lt 1/M$. How can I show the inequality for the $\liminf$ case? I would greatly appreciate any suggestions or solutions.
 A: Let $\liminf s_n=s$, $\liminf\sigma_n=\sigma$ and suppose $\sigma<s$. Pick $t\in (\sigma,s)$. Then for infinitely many $n$, $\sigma_n<t$. Pick $l\in(t,s)$. Then for infinitely many $n$, $s_n<l$ and hence $s\leq l<s$. Contradiction.
OR
For any $\epsilon>0$, for $n$ large enough $s-\epsilon< s_n$. Hence for any $\epsilon_1>0$ for $n$ large enough $s-\epsilon-\epsilon_1<\sigma_n$ . Hence $s-\epsilon-\epsilon_1\leq\sigma$. Since this holds for any $\epsilon$,$\epsilon_1$, $s\leq\sigma$.
A: I've thought of a solution similar to the one used for proving the limsup case, however, I'm not sure if this is correct.
For any $n\gt N$, $\sigma_n=\frac{s_1+s_2+\cdots s_N+ \cdots s_n}{n}\ge \frac{s_1+\cdots +s_N}{n}+\frac{n-N}{n}\dot \inf\{s_n:n\gt N\}$. 
Hence we have $\inf\{\sigma_n: n\gt N\} \ge \frac{s_1+\cdots +s_N}{n}+\frac{n-N}{n}\dot \inf\{s_n:n\gt N\}$. 
Now letting $n\to \infty$, we have $\inf\{\sigma_n: n\gt N\} \ge \inf\{s_n:n\gt N\}$. And then letting $N\to \infty$, we get the desired inequality. 
