Cesaro means $\to \infty$ Let $\{ a_{n}\}_{n}$ be a sequence and let $a\in \mathbb{R}$. Define $\{ c_{n}\}_{n}$ as:

$$c_{n}=\frac{a_{1}+...+a_{n}}{n}.$$

I want to prove the following claim: if $\lim\limits_{n\to +\infty}a_{n}=+\infty$ then $\lim\limits_{n\to +\infty}c_{n}=+\infty$

Approach: Suppose $\lim\limits_{n\to +\infty}a_{n}=+\infty$

Let $M&gt;0$, $\exists\space n_{0}\in\mathbb{N}$ such that $\forall\space n\ge n_{0}, a_{n}&gt;M$

\begin{align}
c_{n} &amp;=\frac{a_{1}+...+a_{n_{0}-1}+a_{n_{0}}+...+a_{n}}{n} \\
&amp;&gt;\frac{a_{1}+...+a_{n_{0}-1}+(n-n_{0})M}{n}\\
&amp;=\frac{a_{1}+...+a_{n_{0}-1}}{n}+\frac{(n-n_{0})M}{n}
\end{align}

So $\lim\limits_{n\to +\infty}c_{n}\ge\lim\limits_{n\to +\infty}\left[\frac{a_{1}+...+a_{n_{0}-1}}{n}+(1-\frac{n_{0}}{n})M\right]=M$

Since this is true $\forall\space M&gt;0$, then $\lim\limits_{n\to +\infty}c_{n}=+\infty$

Is this approach correct?
You were off to a good start. You showed

$$c_{n}&gt;\frac{a_{1}+...+a_{n_{0}-1}}{n}+\frac{(n-(n_{0}-1))M}{n} $$

for $n\ge n_0.$ Thus $\liminf c_n$ is at least the $\liminf$ of the expression on the right. But the expression on the right has a limit, namely $0+M=M,$ and thus its $\liminf$ is the same. So we have shown $\liminf c_n \ge M.$ Since $M&gt;0$ was arbitrary, $\liminf c_n =\infty,$ proving $\lim c_n =\infty.$
$$\lim_{\infty}(1-\frac {n_0}{n})M=M $$

$$\implies \exists n_1 \in \Bbb N \;: $$
$$n&gt;n_1\implies (1-\frac {n_0}{n})M&gt;\frac {M}{2} $$
on the other hand

$$\lim_{\infty}\frac {a_1+...a_{n_0}}{n}=0\implies $$

$$\exists \; n_2\in \Bbb N  \;:$$
$$ n&gt;n_2\implies \frac {a_1+...a_{n_0}}{n}&gt;\frac {-M}{4} $$
thus for $n&gt;\max (n_0,n_1,n_2) $ we have
$$c_n&gt;\frac {M}{4} $$
and this proves that $$\lim_{\infty}c_n=+\infty. $$