How do we get $\frac {a_n}{1+na_n}\ge \frac{\varepsilon}{1+\varepsilon}\frac 1n$?

Question

Suppose $a_n>0$, $s_n=a_1+\ldots+a_n$ and $\sum a_n$ diverges. If $na_n\ge \varepsilon$ for all $n$ for some $\varepsilon>0$, then $\sum\dfrac{a_n}{1+na_n}$ diverges.

I have a solution that says, since $na_n\ge\varepsilon>0$ for all $n$, we see that $\dfrac {a_n}{1+na_n}\ge \dfrac{\varepsilon}{1+\varepsilon}\dfrac 1n$ for all $n$. Hence by comparison theorem original series diverges.

I can't see how $\dfrac {a_n}{1+na_n}\ge \dfrac{\varepsilon}{1+\varepsilon}\dfrac 1n$ is derived.

Answer 1

$n~a_n\geq \epsilon \implies n~a_n+n~\epsilon~a_n\geq \epsilon+n~\epsilon~a_n\implies a_n~n(1+\epsilon)\geq \epsilon(1+n~a_n)\implies \frac{a_n}{1+n~a_n}\geq \frac{\epsilon}{(1+\epsilon)}~\frac{1}{n}$

Answer 2

For all $n$, we have $$\frac{na_n}{1+na_n}=1-\frac{1}{1+na_n}\geq 1-\frac{1}{1+\varepsilon}=\frac{\varepsilon}{1+\varepsilon}.$$