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.
Best Answer
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.