Prove that if $\lim_{n\to+\infty} a_n = +\infty$ then $\lim_{n\to+\infty} \frac{a_1+a_2+...+a_n}{n} = +\infty$ I have to prove a problem, I arrived to a solution but I don't know it is correct.
The problem:
Prove that if
$$\lim_{x\to+\infty} a_n = +\infty$$
Then
$$\lim_{n\to+\infty} \frac{a_1+a_2+...+a_n}{n} = +\infty$$
I thought about doing this similarly to how you prove Ernesto Cesaro's theorem.
By definition
$$\forall M\gt0 \quad \exists n^*: \forall n \gt n^* \quad a_n > M $$
Now I have
$$\frac{a_1+a_2+...+a_{n^*-1}+a_{n^*}+...+a_n}{n}$$
The first part of the fraction can be replaced with
$$S=\frac{a_1+a_2+...+a_{n^*-1}}{n}$$
And then I have all the terms next to $a_{n^*}$ greater than $M$
$$\frac{a_{n^*}+...+a_n}{n} \gt \frac{(n-n^*)M}{n} $$
$$\frac{a_1+a_2+...+a_{n^*-1}+a_{n^*}+...+a_n}{n}\gt S+\frac{(n-n^*)M}{n}$$
Then I choose $M^*$
$$M^*=S+\frac{(n-n^*)M}{n}$$
I have at the end that
$$\frac{a_1+a_2+...+a_n}{n} \gt M^*$$
You can choose whatever $M$ you want, so the problem is proved.
I don't really know if the proof is rigorous, I tried to use that path and I hope that there could be at least something decent in there.
In the case the proof is not valid, I ask you if you can give some tip on where to start to prove it.
 A: You almost ended the problem. You can suppose that all the elements of the sequence are positive (if not it is enough to add some fixed positive constant to every element of the sequence), so $S>0$. So from what you said, for every $M>0$ you have $n^*$ such that for all $n>n^*$,
$\frac{a_1+a_2+...+a_n}{n}\geq S+\frac{(n-n^*)}{n}\cdot M>\frac{(n-n^*)}{n}\cdot M,$
As $\frac{(n-n^*)}{n}$ tends to 1 when $n$ tends to infinity, there will be some $N$ such that if $n>N$, $\frac{(n-n^*)}{n}>\frac{M-1}{M}$. Thus $\frac{a_1+a_2+...+a_N}{N}>\frac{N-n^*}{N}M>M-1$. As you can do this for all $M$, the result follows.
A: The idea in general is kind of ok, but it is not clear enough. You don't want to choose $M^*$, it should be given to you. What you can do is to notice  that from
$$
\frac{a_1+\cdots+a_n}n\geq S_n+ M-\frac{n^*}n\,M,
$$
you obtain
$$
\liminf_{n\to\infty}\frac{a_1+\cdots+a_n}n\geq \liminf_{n\to\infty}S_n+ M-\frac{n^*}n\,M=M.
$$
As you can do this for $M>0$,
$$
\liminf_{n\to\infty}\frac{a_1+\cdots+a_n}n=\infty
$$
and so
$$
\lim_{n\to\infty}\frac{a_1+\cdots+a_n}n=\infty.
$$
A: Good proof,but I think we can make it easier.
for arbitrary $G>0$,exist an $N\in{N^+}$,for every $n>N$,$|a_n|>G$
with this ,we can select some clever $G$,$N_1,N$,and prove it with inequality.
The proof is very close to the proof of Stolz theorem.
