Is it true that $\limsup \phi\le\limsup a_n$, where $\phi=\frac{a_1+…+a_n}n$?

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  • If $\sigma_n=\frac{s_1+s_2+\cdots+s_n}{n}$ then $\operatorname{{lim sup}}\sigma_n \leq \operatorname{lim sup} s_n$

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Solutions Collecting From Web of "Is it true that $\limsup \phi\le\limsup a_n$, where $\phi=\frac{a_1+…+a_n}n$?"

The answer is yes.

Let’s assume that $\limsup a_n =b\neq+\infty$ (The case $b=+\infty$ is easy). Then for $\epsilon>0$ exists $n_0\in\mathbb N$ such that $a_n<b+\epsilon$ for all
$n>n_0$. For any $n>n_0, \ \dfrac{a_1+a_2+\cdots+a_n}{n}<\dfrac{a_1+a_2+\ldots+a_{n_0}}{n}+\dfrac{n-n_0}{n}(b+\epsilon)$.
Now prove that exists $N\in\mathbb N$ such that $\dfrac{a_1+a_2+\cdots+a_n}{n}<b+3\epsilon, \ \forall \ n>N$.
(Hint: If $b+\epsilon>0, \ \dfrac{n-n_0}{n}(b+\epsilon)<b+\epsilon$.

If $b+\epsilon<0$ find $n_1\in\mathbb N$ such that $\dfrac{n-n_0}{n}(b+\epsilon)<b+r\epsilon, \ \forall n\geq n_1$ where $r$ is small enough s.t. $b+r\epsilon<0$ and $1<r<2$.

Use that $\dfrac{a_1+a_2+\ldots+a_{n_0}}{n}\xrightarrow[n\to\infty]{}0$.)