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 $z>0$.

  • $\begingroup$ Do you assume $a_n\ge 0$? (otherwise I do not see what prevents the secong factor from having infinitely many zeroes as $z\to 0+$) $\endgroup$ – fedja Feb 2 at 13:14
  • $\begingroup$ @fedja, the result appears to be straightforward if $a_n$ is of constant sign, even for large enough $n$...so I would assume that $a_n$ oscillates infinitely often. But yes, the RHS of the second inequality should have many cases whereby it has infinitely many zeros as $z\rightarrow 0^+$. Hence I would assume that it is of constant sign as $z\rightarrow 0^+$. See the addendum... $\endgroup$ – user151814 Feb 2 at 14:13
  • $\begingroup$ One more question. People use the sign $\ll$ in two different meanings: "the LHS is bounded by some constant times the (absolute value of) the RHS" and "the LHS is o-small of the RHS". Which one do you mean? $\endgroup$ – fedja Feb 3 at 17:20

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