Question about Answer to limsup of $\sigma_n=\frac{s_1+s_2+\cdots+s_n}{n}$

Question

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?

Answer 1

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.

Answer 2

\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.