Cesaro means $\to \infty$

Question

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 $M>0$, $\exists\space n_{0}\in\mathbb{N}$ such that $\forall\space n\ge n_{0}, a_{n}>M$

\begin{align} c_{n} &=\frac{a_{1}+...+a_{n_{0}-1}+a_{n_{0}}+...+a_{n}}{n} \\ &>\frac{a_{1}+...+a_{n_{0}-1}+(n-n_{0})M}{n}\\ &=\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 $\forall\space M>0$, then $\lim\limits_{n\to +\infty}c_{n}=+\infty$

Is this approach correct?

Answer 1

You were off to a good start. You showed

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

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 $M>0$ was arbitrary, $\liminf c_n =\infty,$ proving $\lim c_n =\infty.$

Answer 2

$$\lim_{\infty}(1-\frac {n_0}{n})M=M $$

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

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

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