Question about Answer to limsup of $\sigma_n=\frac{s_1+s_2+\cdots+s_n}{n}$ here's the relevant question: If $\sigma_n=\frac{s_1+s_2+\cdots+s_n}{n}$ then $\operatorname{{lim sup}}\sigma_n \leq \operatorname{lim sup} s_n$
In the accepted answer, doesn't the last inequality only work if $\sup_{l\geq k}s_l$ is nonnegative?
The "last inequality" I'm referring to is this:
$$\frac 1n\sum_{j=1}^ks_j+\frac{n-k}n\sup_{l\geqslant k}s_l\leqslant  \frac 1n\sum_{j=1}^ks_j+\sup_{l\geqslant k}s_l.$$
I ran into this issue when trying to prove the analagous statement for liminf, because in the case of liminf I could only get a similar inequality if $\inf_{l\geq k}s_l \leq 0$, as follows:
$$\sigma_n=
\frac 1n\sum_{j=1}^ks_j+\frac 1n\sum_{j=k+1}^ns_j
\geqslant \frac 1n\sum_{j=1}^ks_j+\frac{n-k}n\inf_{l\geqslant k}s_l
$$
From here, if $\inf_{l\geq k}s_l \leq 0$ then I could continue and write 
$\geq\frac 1n\sum_{j=1}^ks_j+\inf_{l\geqslant k}s_l$.
Could someone clarify please?
 A: You have that
$$ \tag{*}
\sigma_n\geqslant \frac 1n\sum_{j=1}^ks_j+\frac{n-k}n\inf_{l\geqslant k}s_l
$$
and you are right that this is  $\ge \frac 1n\sum_{j=1}^ks_j+\inf_{l\geqslant k}s_l$ only if $\inf_{l\geqslant k}s_l \le  0$.
But that estimate is actually not needed: For fixed $k$  you can take the $\liminf_{n \to \infty}$ in $(*)$, this gives
$$
\liminf_{n \to \infty}\sigma_n\geqslant \inf_{l\geqslant k}s_l
$$
because the right-hand side has a limit for $n \to \infty$.
Then take the limit for $k \to \infty$ and conclude that
$$
\liminf_{n \to \infty}\sigma_n\geqslant\liminf_{n \to \infty}s_n\, .
$$
The same approach works for $\limsup$ in the referenced Q&A.
A: \begin{align*}
&\limsup_{n}\left(\dfrac{1}{n}\sum_{j=1}^{k}s_{j}+\dfrac{n-k}{n}\sup_{l\geq k}s_{l}\right)\\
&\leq\limsup_{n}\dfrac{1}{n}\sum_{j=1}^{k}s_{j}+\limsup_{n}\dfrac{n-k}{n}\sup_{l\geq k}s_{l}\\
&=\lim_{n}\dfrac{1}{n}\sum_{j=1}^{k}s_{j}+\lim_{n}\dfrac{n-k}{n}\sup_{l\geq k}s_{l}\\
&=\sup_{l\geq k}s_{l},
\end{align*}
you still got it.
