How to prove an average inequality about limsup\liminf Suppose $an$ is an infinite sequence in $\Bbb R$ and $bn$ is the arithmetic average of $an$, or $bk = \frac{(a_1 + ... + a_k)}k$
Why is it true that:
$\lim$(inf $an) \le$ $\lim$(inf $bn) \le \lim$(sup $bn) \le \lim$(sup $an$)
 A: For ease of notation let $$
\ell_n = \inf_{n\leq m < \infty}a_m
$$
For the first inequality use the $\epsilon-\delta$ definition of $\inf$: For any $\epsilon > 0$ there exists some $N$ (possibly depending on $n$) so that $\ell_n < a_m+\epsilon $ for all $m > N$, hence
\begin{align}
\ell_n = \frac{(m-N)\ell_n}{m-N}
%
&< \frac{(a_{N+1}+\epsilon)+\dotsb+ (a_m+\epsilon) }{m-N}\\
%
&= \frac{a_{N+1}+\dotsb+ a_m}{m}\cdot \frac{m}{m-N} + \epsilon\\
%
&= \left(b_n - \frac{a_1+\dotsb+ a_N}{m}\right) \cdot \frac{m}{m-N} + \epsilon
\end{align}
Given that it holds for all $m > N$, 
\begin{align}
\ell_n
&\leq \lim_{m\to\infty}\bigg(b_n - \underbrace{(a_{1}+\dotsb+ a_N)}_{\text{constant in $m$}}\cdot\underbrace{\frac{1}{m}}_{\to\,0}\bigg)\cdot \underbrace{\frac{m}{m-N}}_{\to\,1} + \epsilon
= b_n+\epsilon
\end{align}
This holds for all $\epsilon > 0$, so let $\epsilon \to 0^+$, giving
$\ell_n \leq b_n$. The first inequality follows:
\begin{align}\operatorname{lim\,inf}_{n\to\infty}a_n
&= \operatorname{lim\,inf}_{n\to\infty}\inf_{n\leq m < \infty}a_m
%
= \operatorname{lim\,inf}_{n\to\infty} \ell_n\\
%
&\leq \operatorname{lim\,inf}_{n\to\infty} b_n
\end{align}
For the second inequality, trivially
$$\operatorname{lim\,inf}_{n\to\infty} b_n \leq \operatorname{lim\,sup}_{n\to\infty} b_n$$
And finally, for the third inequality consider $(-a_n)$ and $(-b_n)$. Applying the first inequality to these (noting that the arithmetic mean of negation is the negation of the arithmetic mean) we have
$$\operatorname{lim\,inf}_{n\to\infty} (-a_n)
\leq \operatorname{lim\,inf}_{n\to\infty} (-b_n)$$
and given that $\operatorname{lim\,sup}_{n\to\infty}(-x_n) = -\operatorname{lim\,inf}_{n\to\infty}x_n$
we conclude
$$-\operatorname{lim\,sup}_{n\to\infty}a_n
\leq -\operatorname{lim\,sup}_{n\to\infty} b_n$$
