Proving if $\lim_{n\rightarrow\infty}a_n=L $ then $\lim_{n\rightarrow\infty} \frac{a_1+a_2+\cdots+a_n}n=L $ [duplicate]

Good morning, i have a big problem with this proof. I haven't idea about this proof.

Problem:

Suppose $\lim_{n\rightarrow\infty}a_n =L $ then $\lim_{n\rightarrow\infty}\ \frac{a_1+a_2+\cdots+a_n}n=L $

I've tried this:

$| a_n-L|<\varepsilon\Rightarrow L-\varepsilon<a_n<L+\varepsilon$ Please help!!

Let $\epsilon>0$. Therefore, there exists $N \in \mathbb{N}$ such that $m> N \implies a_m < L+\epsilon$. Therefore,

$$\frac{a_1+\cdots+a_n}{n}=\frac{a_1+\cdots +a_N}{n}+\frac{a_{N+1}+\cdots+a_n}{n}$$ $$< \frac{a_1+\cdots +a_N}{n}+\frac{(n-(N+1))(L+\epsilon)}{n}.$$

Passing the $\limsup$ as $n \to \infty$, we get

$$\limsup\limits_{n \to \infty} \frac{a_1+\cdots+a_n}{n} \leq L+\epsilon.$$ Analogously, we can prove $$\liminf\limits_{n \to \infty} \frac{a_1+\cdots+a_n}{n} \geq L-\epsilon.$$ But this holds for every $\epsilon>0.$ Therefore, $$\lim\limits_{n \to \infty} \frac{a_1+\cdots+a_n}{n}=L.$$

Here's a version of the proof that maybe looks a little longer than other versions, but which has the advantage that each step is more or less obvious:

First, we introduce the notation $$\sigma_n=\frac{a_1+\dots+a_n}{n}.$$

Lemma If there exists $N$ so that $a_n=0$ for all $n>N$ then $\lim\sigma_n=0$.

Proof: $$|\sigma_n|\le\frac{|a_1+\dots+a_N|}{n}.$$QED

Proposition If $\lim a_n=0$ then $\lim\sigma_n=0$.

Proof: Let $\epsilon>0$. Choose $N$ so $|a_n|<\epsilon$ for all $n>N$. Define $$a_n'=\begin{cases}0,&(n\le N), \\a_n,&(n>N).\end{cases}$$The lemma shows that $$\lim(\sigma_n-\sigma_n')=0$$(using what should be obvious notation). But $|a_n'|\le\epsilon$ for every $n$, so $|\sigma_n'|<\epsilon$ for every $n$, hence $$\limsup|\sigma_n'|\le\epsilon.$$Since $\sigma_n-\sigma_n'\to0$ this shows that $\limsup|\sigma_n|\le\epsilon$, and since this holds for every $\epsilon>0$ it follows that $\limsup|\sigma_n|=0$, hence $\lim\sigma_n=0$. QED.

Theorem If $\lim a_n=L$ then $\lim \sigma_n=L$.

Proof: Let $a_n'=a_n-L$. Then the lemma shows that $\lim\sigma_n'=0$, since $\lim a_n'=0$. But $\sigma_n=\sigma_n'+L$, hence $\lim \sigma_n=L$. QED