Please prove divergence of arithmetic mean.

My textbook has the following proposition in it.
However, I only know how to solve the problem halfway.

Proposition
$a_n \to \infty (n \to \infty) \Rightarrow b_n=\frac{a_1+\dots +a_n}{n}\to \infty$

My proof (halfway through)
For any $K \in \mathbb{R}$ there exists a number $N \in \mathbb{N}$, $n> N\Rightarrow a_n > K$ .

$b_n = \frac{a_1+\dots+a_n}{n}=\frac{a_1+\dots+a_N}{n}+\frac{a_N+1+\dots+a_n}{n}$

$>\frac{M}{n}+\frac{n-N}{n}K$ $(M:=a_1+\dots+a_N)$

$=\frac{M}{n}+(1-\frac{N}{n})K$

$=K+\frac{1}{n}(M-NK)$

$=\dots$

What should I do after this?
Please tell me the rest of the proof.

Solution 1:

For any $K \in \mathbb{R}^+$ there exists a number $N_1 \in \mathbb{N}$ such that $$n> N_1\Rightarrow a_n > 2K.$$ Now fix $N_1$ and then for $n> N_1$, \begin{eqnarray} b_n &=& \frac{a_1+\dots+a_n}{n}=\frac{a_1+\dots+a_{N_1}}{n}+\frac{a_{N_1+1}+\dots+a_n}{n}\\ &\ge&\frac{M}{n}+\frac{n-N_1}{n}\cdot2K \\ &=&2K+\frac{1}{n}(M-2N_1K) \end{eqnarray} where $$ M:=a_1+\dots+a_{N_1}. $$ Since $M, N_1, K$ are finite numbers, there is $N_2\in\mathbb{N}$ such that $$n> N_2\Rightarrow K+\frac{1}{n}(M-2N_1K)>0.$$ Define $N=\max\{N_1,N_2\}$. Then for given $K>0$ and above $N$, one has $$ n>N\Rightarrow b_n>K. $$