& & \le -\frac{1}{2}\sum_{k=1}^\infty b_k \log{(1+4e^{-\alpha_n k}\sin^2{(\pi
  uk)})} \nonumber \\
  & & \le -\frac{\log{5}}{2}\sum_{k=1}^\infty b_k e^{-\alpha_n
  k}\sin^2{(\pi uk)} \nonumber \\
  & & =-\frac{\log{5}}{2} S_n \le -\frac{\log{5}}{2}
  C_2\alpha_n^{-\epsilon},
 \end{aligned}$$ where the last two inequalities follow from the fact that $\log{(1+y)}\ge\left(\frac{\log{5}}{4}\right)y$ $(0\le y\le 4)$ and (\[sinus\]), respectively. Thus, the required inequality is proved for $\alpha_n\le\mid\theta\mid<\pi$. It remains to consider the interval $\delta_n\le\mid\theta\mid<\alpha_n$. (\[regtau\]) implies that we have to find now a lower bound for the sum $S_n$ in (\[sinus\]) if $\delta_n/2\pi\le\mid u\mid<\alpha_n/2\pi$ (i.e. if $\delta_n\le\mid\theta\mid<\alpha_n$). We shall apply Wiener-Ikehara Tauberian theorem setting there $c_k=k^{-\rho+1}b_k, k\ge 1$, and $$\label{dtil}
 \tilde{D}(s)=\sum_{k\ge 1}
 b_k k^{-s-\rho+1}=D(s+\rho-1), \quad s=\sigma+iy.$$ Since both $C_0,\rho>0$ from condition $(M_1)$, the function $\tilde{D}(s)$ satisfies the condition of Wiener-Ikehara theorem with $c=A$. Moreover, since by (\[alphan\]) $\alpha_n^{-1}\to\infty$ as $n\to\infty$, we can apply (\[wi\]) in the form $$\label{wik}
 \sum_{1\le k\le K^/\alpha_n} k^{-\rho+1}b_k\sim AK/\alpha_n,
 \quad n\to\infty,$$ where the constant $K>0$ will be specified later. Our next argument will be similar to that given in \[8; Lemma 7\]. First, using (\[gam\]), we observe that $\parallel uk\parallel=uk$ if $\mid
 u\mid k<1/2$. This implies that, for $1\le k\le\pi/\alpha_n$, $\parallel
 uk\parallel$ can be replaced by $\mid u\mid k$. Recalling that $\mid u\mid\ge\delta_n/2\pi$ and applying (\[sinlowbound\]) and (\[delta\]), we obtain 