inequality involving two dirichlet series

Question

Let $f\left( s \right) = \sum\limits_{n = 1}^{\infty}\left[ a_{n} \cdot \left( {\frac{1}{n^{s}}-\frac{1}{n^{1 - \operatorname{conj}\left( s \right)}}} \right) \right]$ and Let $g\left( s \right) = \sum\limits_{n = 1}^{\infty}\left[ c_{n} \cdot \left( {\frac{1}{n^{s}} - \frac{1}{n^{1 - \operatorname{conj}\left( s \right)}}} \right) \right]$ be two Dirichlet series with different intervals of convergence. the question is: if $\left| \left( \frac{a_{n}}{n^{s}} \right) \right| \le \left| \left( \frac{c_{n}}{n^{s}} \right) \right|$ does this means that $\left| f\left( s \right) \right| \le \left| g\left( s \right) \right|$ ?

Answer 1

It would be great but unfotunately, it is not true.

Let's consider $a_n=n^s(-2t)^n/n!$ and $b_n=n^st^n/n !$

Then you have $|a_n|/n^s>|b_n|/n^s$ but you have also :

$$\sum_n a_n/n^s = e^{-2t}$$ And $$\sum_n b_n/n^s = e^{t}. $$