& =\frac{1}{\sqrt{\mathcal{B}_b(e^{-\alpha_n})}}, \quad
 n\to\infty. \nonumber
 \end{aligned}$$ The last asymptotic equivalence follows from (\[balpha\]) of Lemma 2, which implies that $$\delta_n\sqrt{\mathcal{B}_b(e^{-\alpha_n})} \sim
\frac{\alpha_n^{-\rho/6}}{\omega(n)} \sqrt{h(\rho+1)}\to\infty,$$ if $\omega(n)\to\infty$ slower than $\alpha_n^{-\rho/6}$.

Substituting the asymptotic equivalence for $I_n$ into (\[jonei\]), we conclude that $$\label{joneas}
J_1(m,n)\sim
\frac{e^{n\alpha_n}}{\sqrt{2\pi\mathcal{B}(e^{-\alpha_n})}}
e^{F_{m,b}(e^{-\alpha_n})}$$ if $m$ satisfies (\[mn\]) as $n\to\infty$.

For the estimate of $J_2(m,n)$, we recall (\[jtwo\]) and the proof of (\[realparts\]). Thus, for any real $u$, we obtain $$\begin{aligned}
 \label{realpartsfm}
& & \Re{(F_{m,b}(e^{-\alpha_n+2\pi iu}))} -F_{m,b}(e^{-\alpha_n})
\le -\frac{\log{5}}{2}\sum_{k=1}^m b_k e^{-\alpha_n k}\sin^2{(\pi
uk)} \nonumber \\
& & =-\frac{\log{5}}{2}\left(\sum_{k=1}^\infty b_k e^{-\alpha_n
k}\sin^2{(\pi uk)} -\sum_{k=m+1}^\infty b_k e^{-\alpha_n
k}\sin^2{(\pi uk)}\right).\end{aligned}$$ Consider again the sequences $m=m(n)$ and $m_1=m_1(n)$ defined by (\[mn\]) and (\[monen\]), respectively. We have $$\begin{aligned}
\label{twosums}
 & & \sum_{k=m+1}^\infty b_k e^{-\alpha_n k} \sin^2{(\pi ku)}
\le\sum_{k=m+1}^\infty b_k e^{-\alpha_n k} \nonumber \\
 & & =\sum_{k=m+1}^{m_1}b_k
e^{-\alpha_n k} +\sum_{k=m_1+1}^\infty b_k e^{-\alpha_n k}.
 \end{aligned}$$ These sums can be estimated using the argument given in the proof of Lemma 7 (see (\[sigmaprev\])-(\[stwo\])). We obtain in the same way that $$\label{firsts}
\sum_{k=m+1}^{m_1}b_k e^{-\alpha_n k}
=O\left(\alpha_n^\rho\sum_{k=m+1}^{m_1} b_k\right) =O((\alpha_n
m_1)^\rho) =O(\log^\rho{\alpha_n^{-1}}),$$ $$\label{seconds}
\sum_{k=m_1+1}^\infty b_k e^{-\alpha_n k}
=O\left(\sum_{k=m_1+1}^\infty k^\rho e^{-\alpha_n k}\right)
=O(\log^\rho{\alpha_n^{-1}})$$ and thus, (\[twosums\])-(\[seconds\]) imply that $$\sum_{k=m+1}^\infty b_k e^{-\alpha_n k} \sin^2{(\pi ku)}
=O(\log^\rho{\alpha_n^{-1}}).$$ Replacing the second term of the right-hand side 