proving that if $lim_{n\to\infty}a_n=\infty$ then $lim_{n\to\infty}b_n=\infty$, where $b_n=\frac{1}{n}\sum_{i=1}^n{a_i}$.

I have to prove that if $lim_{n\to\infty}a_n=\infty$ then $lim_{n\to\infty}b_n=\infty$, where $b_n=\frac{1}{n}\sum_{i=1}^n{a_i}$.

What I've got:

Let $\epsilon > 0$. We know that $a_n\to \infty$, so we fix $n_0$ s.t. when $n > n_0$ we have $a_n > \frac{\epsilon}{2}$. We fix $k \in \mathbf N, k > n_0$. Let $n > k$, so we have $$b_n > \frac{1}{n}\sum_{i=1}^{k-1}a_i + \frac{n-k+1}{2n}\epsilon$$

That's where I got stuck. Basically I want to express $\frac{1}{n}\sum_{i=1}^{k-1}a_i$ somehow in terms of $\frac{\epsilon}{2}$.

$a_n \to \infty$ means: for all $M > 0$ there is $N > 0$ such that $a_n > M$ whenever $n > N$. Moreover, we can choose $N$ to be large enough that $\sum_{i=1}^n a_i > 0$ for $n > N$, because $a_i \to \infty$.

Then, if $n > 2N$, $$b_{n} = \frac{1}{n} \sum_{i=1}^{n} a_i = \frac{1}{n} \sum_{i=1}^N a_i + \frac{1}{n} \sum_{i=N+1}^{n} a_i.$$ Because $a_i > M$ for $i > N$ and $N/n < 1/2$, the second sum on the RHS is at least $$\frac{1}{n} \cdot (n-N) \cdot M = M - M \cdot (N/n) > M/2.$$ And we chose $N$ large enough that the first sum on the RHS is positive. So the total sum is at least $M/2$.

That is, if $n > 2N$, then $b_n > M/2$. But $M$ was arbitrary, so this means $b_n \to \infty$.