\[12; Thm. 2.2, p. 122\].) Suppose that the Dirichlet series $\tilde{D}(s)=\sum_{k=1}^\infty
 c_k k^{-s}$ is such that the function $\tilde{D}(s)-\frac{c}{s-1}$ has an analytic continuation to the closed half-plane $\Re(s)\ge 1$. Then $$\label{wi}
 \sum_{k=1}^n c_k\sim cn, \quad n\to\infty. \quad \rule{2mm}{2mm}$$*]{}

We also denote by $\{\gamma\}$ the fractional part of the real number $\gamma$, and by $\parallel\gamma\parallel$ the distance from $\gamma$ to the nearest integer, so that $$\label{gam}
 \parallel\gamma\parallel=\left\{\begin{array}{ll} \{\gamma\} & \qquad  \mbox {if}\qquad \{\gamma\}\le 1/2, \\
 1-\{\gamma\} & \qquad \mbox {if}\qquad \{\gamma\}>1/2.
 \end{array}\right.$$ It is not difficult to show that $$\label{sinlowbound}
 \sin^2{(\pi\gamma)}\ge 4\parallel\gamma\parallel^2$$ (see \[8; p. 272\]). Now, we are ready to prove that Hayman last (“decay”) condition \[7; p. 565\] is also valid.

Suppose that $f_b(x)$ satisfies conditions $(M_1)-(M_3)$. Then, for sufficiently large $n$, $$\mid f_b(e^{-\alpha_n+i\theta})\mid\le f_b(e^{-\alpha_n})
 e^{-C_3\alpha_n^{-\epsilon_1}}$$ uniformly for $\delta_n\le\mid\theta\mid<\pi$, where $C_3$ and $\epsilon_1$ are positive constants.

[*Proof.*]{} First, we notice that $$\label{modul}
\frac{\mid f_b(e^{-\alpha_n+i\theta})\mid}{f_b(e^{-\alpha_n})}
=\exp{(\Re({\log{f_b(e^{-\alpha_n+i\theta}))}}
-\log{f_b(e^{-\alpha_n})})}.$$ Then, setting $\theta=2\pi u$, for $\alpha_n/2\pi\le\mid u\mid=\mid\theta\mid/2\pi<1/2$, we almost repeat the argument from \[10; p. 324\]: $$\begin{aligned}
 \label{realparts}
  & & \Re{(\log{f_b(e^{-\alpha_n+i\theta})})}
-\log{f_b(e^{-\alpha_n})} \nonumber \\
 & & =\Re{\left(-\sum_{k=1}^\infty b_k \log{\left(\frac{1-e^{-k\alpha_n+2\pi
 iuk}}{1-e^{-k\alpha_n}}\right)}\right)} \nonumber \\
  & & =-\frac{1}{2}\sum_{k=1}^\infty b_k
  \log{\left(\frac{1-2e^{-k\alpha_n}\cos{(2\pi uk)}+e^{-2\alpha_n k}}
  {(1-e^{-\alpha_n k})^2}\right)} \nonumber \\
  & & =-\frac{1}{2}\sum_{k=1}^\infty b_k
  \log{\left(1+\frac{4e^{-\alpha_n k}\sin^2{(\pi uk)}}{(1-e^{-\alpha_n
  k})^2}\right)} \nonumber \\
  