Suppose $a_{n} = O(1)$ and

$$f(z)=\sum_{n=1}^{\infty} \frac{n a_{n}e^{-nz}}{1-e^{-nz}} \ll z^{-1}$$ as $z\rightarrow 0^+$. Is it true that

$$f(z) \ll \frac{1}{z} \Bigg(\sum_{n=1}^{\infty} \frac{a_{n}e^{-nz}}{1-e^{-nz}}\Bigg)$$ as $z \rightarrow 0^+$ ?

*Addendum:* As @fedja commented below, it appears the RHS of the second inequality could have infinitely many zeros as $z \rightarrow 0^+$. Hence it may be assumed that $\sum_{n=1}^{\infty} \frac{a_n e^{-nz}}{1-e^{-nz}} $ is of constant sign for small $z>0$.