Prove : if a real sequence diverges to infinity , then its arithmetic mean sequence diverges to infinity

Question : if $$\lim_{n\to \infty} a_n = \infty$$ then $$\lim_{n\to \infty} \frac 1n\sum_{k=1}^n a_k = \infty$$

My attempt for solving : $$\lim_{n\to \infty} a_n = \infty \Rightarrow \forall M \exists N\in \Bbb N : \forall n > N \Rightarrow a_n >M$$

To prove by definition we need to show that $$\forall M \exists n_0\in \Bbb N : \forall n > n_0 \Rightarrow \frac 1n\sum_{k=1}^n a_k >M$$

Now I am trying to find $$n_0$$ to satisfy the conditions and write in the formal proof :

Let us define $$a_m = min \{ a_1 , a_2 , \ldots ,a_N\}$$

$$\frac 1n\sum_{k=1}^n a_k = \frac 1n\sum_{k=1}^N a_k + \frac 1n\sum_{k=N+1}^n a_k >\frac 1nNa_m + \frac 1n\left( n - N \right)M$$

From here on I got stuck and couldn't figure how to find my $$n_0$$ .

Another thought : I tried using a previous proof saying : if $$\lim_{n\to \infty} a_n = \infty$$ and for each $$n\in \Bbb N : a_n \le b_n$$ Then it follows that $$\lim_{n\to \infty} b_n = \infty$$ .

I defined the AM sequence as $$b_n$$ but I failed to find a sequence $$a_n$$ that satisfies these conditions .

First, you can take $$n = n(M)$$ large enough, so that
$$\frac{1}{n}\sum_{k=1}^na_k > \frac{n-N}{n}M = \big(1-\frac{N}{n}\big)M > \frac{M}{2}$$
and you can take $$M$$ as large as you want
• I assumed that $a_k > 0$; otherwise the statement is wrong. The counterexample: a sequence: $$1, 1, -2, 2, 2, -4, \ldots, 2^k, 2^k, -2^{k+1}, \ldots$$ Every sum of first $3k$ terms is 0. The first $>$ sign follows from your inequality $\frac{1}{n}\sum_{k=1}^na_k > \frac{1}{n}Na_m + \frac{1}{n}(n-N)M$ – diplodoc Jul 27 at 9:49