How do we prove $\lim\inf s_n\le\lim\inf\sigma_n$ in the following case? Let $s_n$ be a sequence of nonnegative numbers, and for each $n$ define $\sigma_n=\frac1n (s_1+s_2+...+s_n)$, how do we prove  $\lim\inf s_n\le\lim\inf\sigma_n$?
I was told two ways could be achieved:
$1.$ Suppose we have proved $\lim\sup \sigma_n\le\lim\sup s_n$, then we can do a little trick to prove the $\lim\inf$ version. My thought is that if we can somehow make $\lim\sup -\sigma_n\le\lim\sup -s_n$, then we can do $\lim\inf\sigma_n=-\lim\sup -\sigma_n\ge-\lim\sup -s_n=\lim\inf s_n$, but the problem is how do we make $\lim\sup -\sigma_n\le\lim\sup -s_n$.
$2.$ Directly prove it. My thought is that we can pick a $N$, $n>N\implies$ $\sigma_n=\frac1n (s_1+s_2+...+s_n)\ge \min\{s_1, ...,s_n\}\ge\inf s_n$. Since $\inf s_n$ is lower bound for $\sigma_n$, $\inf s_n\le\inf\sigma_n$, then taking the limit, we get $\lim\inf s_n\le\lim\inf\sigma_n$. But I felt taking limit part is not quite logical to me.
Could someone help?
 A: Fix $n\in\mathbb{N}$. Now, let $k\geq n$ be arbitrary. Then,
$$
\begin{aligned}
\sigma_{k}&=\frac{1}{k}\sum_{i=1}^{k}s_{i}\\
&=\frac{1}{k}\left(\sum_{i=1}^{n}s_{i}+\sum_{i=n+1}^{k}s_{i}\right)\\
&=\frac{1}{k}\left(\sigma_{n}+\sum_{i=n+1}^{k}s_{i}\right)\\
&\geq\frac{1}{k}\left(\sigma_{n}+(k-n-1)\inf_{i\geq n+1}s_{i} \right)
\end{aligned}
$$
Since this holds for all $k\geq n$ and using the fact that $\inf(a_{n}+b_{n})\geq\inf(a_{n})+\inf(b_{n})$, it follows that
$$
\inf_{k\geq n}\sigma_{k}\geq\inf_{k\geq n}\frac{\sigma_{n}}{k}+\inf_{k\geq n}\frac{k-n-1}{k}\inf_{i\geq n}s_{i}.
$$
Taking $n\to\infty$ gives
$$
\liminf\sigma_{n}\geq\liminf s_{n}.
$$
A: Fix a $k\leq n$, 
$\sigma_n=\frac 1 n \sum _1 ^{k-1} s_i + \frac 1 n \sum _k ^n s_i$
$\geq\frac 1 n \sum _1 ^{k-1} s_i + \min \{s_k,s_{k+1},..s_n \}$
$\geq\frac 1 n \sum _1 ^{k-1} s_i + \inf_{i\geq k} s_i$
Note that this holds for all $n\geq k$. For all finite $k$,
$\inf_{j\geq n} \sigma_j\geq \frac 1 n \sum _1 ^{k-1} s_i + \inf_{i\geq k} s_i$
Now we first take the limit w.r.t $n$ on both sides, and note that for all finite $k$,
$\liminf \sigma_j\geq \inf_{i\geq k} s_i$
Now, we take limit w.r.t $k$ on both sides, to get
$\liminf \sigma_n\geq \liminf s_n$
A: Your idea is correct and you have to work a little bit harder.
We can write $\liminf _n s_n = \sup _{k}\inf_{n\geq k} s_n $.
We now fix $k$, and denote $  S_n= \sum _{i=1}^{n}s_n      $
So, 
\begin{align}
  \sigma _n&= \frac{   S_{k-1}   }{n}  +  \frac{ S_n-  S_{k-1}   }{n}\\
&\geq \frac{   S_{k-1}   }{n}  +  \frac{ (n-k)\min_{n\geq i\geq k}s_i   }{n}\\
&\geq \frac{   S_{k-1}   }{n}  +  \frac{ (n-k)\inf_{ i\geq k}s_i   }{n}
\end{align}
Taking $\liminf_n$ in both sides we obtain (notice that the limit of the RHS exists),
$$\liminf _n \sigma_n \geq \inf_{i\geq k}s_i$$
Finally take $\sup$ with the respect of $k$ of the previous inequality we deduce
$$\liminf _n \sigma_n \geq \sup _{k}\inf_{i\geq k}s_i =\liminf _is_i$$
A: You can almost immediately see that
$ \min\{s_1, ...,s_n\} \le  \min\{\sigma_1, ...,\sigma_n\} \le \max\{\sigma_1, ...,\sigma_n\} \le  \max\{s_1, ...,s_n\}$
So,
$\liminf s_{n}\leq\liminf \sigma_{n} \le \limsup\sigma_{n}\leq\limsup s_{n}$
