Show that if $a_n> 0$ and $\lim na_n = \ell$ with $\ell \neq 0$ then the series $\sum a_n$ diverges I'm trying to understand the solution to the following problem:

Show that if $a_n> 0$ and $\lim na_n = \ell$ with $\ell \neq 0$ then the series $\sum a_n$ diverges.

I don't understand how the sign changed in the starred equation. Can someone briefly explain the solution?
If $na_n \to 1$ and $a_n > 0$ then $l \geq 0$. Furthermore, we know that for any $\epsilon>0$ there is an $N$ for which
$n > N$ implies $|na_n-\ell| < \epsilon$. Equivalently, $|a_n-\ell/n| < \epsilon/n$. But then
$$ a_n > \frac{\ell}{n} - \frac{\epsilon}{n}. \tag{$\ast$} $$
Consider this for $\epsilon = \ell/2 > 0$. Then for $n > N$ we have $a_n > \ell/2n$. But $\ell/2 \sum 1/n$ diverges. Hence $\sum a_n$ also diverges.

The claim $|a_n - \frac{\ell}{n}| < \frac{\epsilon}{n}$ is equivalent to
$$ -\frac{\epsilon}{n} < a_n - \frac{\ell}{n} < \frac{\epsilon}{n}. $$
In particular, the left inequality implies $(\ast)$.
Why is $|X| < Y$ equivalent to $-Y < X < Y$? By definition, $|X| = \max(X,-X)$. Now $\max(X,-X) < Y$ iff both $X < Y$ and $-X < Y$. The latter is equivalent to $X > -Y$, and put together we get $-Y < X < Y$.
