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 .
Best Answer
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