Q:

Basic analysis question: $\max_{1\leq i\leq n} a_i\geq n\epsilon\implies a_n\geq n\epsilon$

This is a follow up to something I asked earlier. (This question is self-contained so you don't need to click the link.) Thank you very much for your help!

Question: $\{a_n\}$ is a sequence of nonnegative real numbers and $\epsilon>0$
  is given. We know that there are arbitrarily large $n$'s satisfying
  $$
 \max_{1\leq i\leq n}a_i\geq n\epsilon.\tag{A} 
$$ 
  How can I show this: there are arbitrarily large $n$'s such that
  $$ a_n\geq n\epsilon.\tag{B} 
$$

Attempt: I have started by arguing that if $a_n<n\epsilon$ for all $n$, then for any $n$, 
$$
\max_{1\leq i\leq n}a_i<\max\{\epsilon,2\epsilon,\ldots,n\epsilon\}=n\epsilon
$$ 
so (A) is violated. It follows that there is $n_1\geq 1$ such that $a_{n_1}\geq n_1\epsilon$. Next, because of (A), there is some $N$ such that $N\epsilon>a_{n_1}$ and
$$
\max_{1\leq i\leq N}a_i\geq N\epsilon.
$$
How do I continue from here?

A:

Fix an $N$ such that $\max_{1 \leq i \leq N} a_i \geq N\epsilon$. Then, there exists ${m_1}$ with $1 \leq m_1 \leq N$ such that $a_{m_1} = \max_{1 \leq i \leq N} a_i$. Note that $a_{m_1} \geq N\epsilon \geq m\epsilon$.
Since there are arbitrarily many $n$ that satisfy (A), consider an $n$ with $n > \displaystyle\frac{a_{m_1}}{\epsilon}$such that $\max_{1 \leq i \leq n} a_i \geq n\epsilon$. Repeating as above, we can find $m_2$ such that $a_{m_2} \geq m_2\epsilon$.
The reason for picking $n > \displaystyle\frac{a_{m_1}}{\epsilon}$ is so that $m_1$ does not satisfy $a_i \geq n\epsilon$. If $m_2 \leq N$, then $m_2$ would have been chosen instead of $m_1$ in the first step. Hence, we can conclude that $m_2 > m_1$. Also, when picking the $m_i$, we can say that we're picking the largest possible index that achieves the maximum over $1 \leq i \leq N_i$ so we can have $m_1 \leq N_1 < m_2 \leq N_2 \cdots$.
Repeating the above process, we see that there are arbitrarily many $n$ such that $a_n \geq n\epsilon$.

