u^\rho du =O(m_1^{\rho+1}).$$ Therefore $$\label{secsum}
 O\left(\alpha_n^\rho\sum_{k=m+1}^{m_1}k^\rho\right)
=O(\alpha_n^\rho m_1^{\rho+1})
=O(\alpha_n^{-1}\log^{\rho+1}{\alpha_n^{-1}}).$$ The first sum in (\[sigmaprev\]) can be also rewritten as $$\begin{aligned}
\label{simplif}
& & \sum_{k=m+1}^{m_1}k^{\rho+1}(L_k-L_{k-1}) =-L_m
(m+1)^{\rho+1}+L_{m_1}m_1^{\rho+1} \nonumber \\
& & +\sum_{j=1}^{m_1-m-1}L_{m+j}((m+j)^{\rho+1}
-(m+j+1)^{\rho+1}).\end{aligned}$$ We recall that by Lemma 6, (\[mn\]) and (\[monen\]), $L_m=o(1)$ and $L_{m_1}=o(1)$. Hence $$\label{t}
(-L_m(m+1)^{\rho+1}+L_{m_1}m_1^{\rho+1})\alpha_n^\rho
=o(\alpha_n^{-1}\log^{\rho+1}{\alpha_n^{-1}}).$$ The last sum in (\[simplif\]) is estimated using again an approximation by an integral. First, applying a binomial expansion, we get $$\begin{aligned}
& & (m+j)^{\rho+1}-(m+j+1)^{\rho+1}
=(m+j)^{\rho+1}\left(1-\left(1+\frac{1}{m+j}\right)^{\rho+1}\right)
\nonumber \\
& & =(m+j)^{\rho+1}\left(-\frac{\rho+1}{m+j}+O((m+j)^{-2})\right)
\nonumber \\
& & =-(\rho+1)(m+j)^\rho+O((m+j)^{\rho-1}). \nonumber\end{aligned}$$ Then, from (\[mn\]), (\[monen\]) and the fact that $L_{m+j}$ are bounded for $1\le j\le m_1-m-1$ we obtain $$\begin{aligned}
\label{bigsum}
& &\sum_{j=1}^{m_1-m-1}L_{m+j}((m+j)^{\rho+1}-(m+j+1)^{\rho+1})
\nonumber\\
& & =-(\rho+1)\sum_{j=1}^{m_1-m-1}
L_{m+j}((m+j)^\rho+O((m+j)^{\rho-1})) \nonumber \\
& & =-(\rho+1)
m_1^{\rho+1}\sum_{\frac{1}{m_1}\le\frac{j}{m_1}<1-\frac{m}{m_1}}
L_{m+j}\left(\frac{1}{m_1}\right)
\left(\left(\frac{m+j}{m_1}\right)^\rho
+O\left(\frac{1}{m_1}\left(\frac{m+j}{m_1}\right)^{\rho-1}\right)\right)
\nonumber \\
& & =O\left(m_1^{\rho+1} \int_0^{1-\frac{\rho}{\rho+1}}
\left(\left(\frac{\rho}{\rho+1}+u\right)^\rho
+O\left(\frac{1}{m_1}\right)\right)du\right) =O(m_1^{\rho+1}).\end{aligned}$$ Combining (\[monen\]) with (\[sigmaprev\])-(\[bigsum\]), we obtain $$\label{sone}
\Sigma_1=O(\alpha_n^{-1}\log^{\rho+1}{\alpha_n^{-1}}).$$

$\Sigma_2$ can be estimated in a similar way. We have $$\begin{aligned}
 \label{stwo}
 & & \Sigma_2
 =O\left(\sum_{k=m_1+1}^\infty
 \frac{k^{\rho+1}e^{-k\alpha_n}}{1-e^{-k\alpha_n}}\right) =O\left(\sum_{k=m_1+1}^\infty
k^{\rho+1}e^{-k\alpha_n}\right) \nonumber \\
& & =O\left(\alpha_n^{-\rho-2}\int_{m_1\alpha_n}^\infty
u^{\rho+1}e^{-u}du\right)
=O(\alpha_n^{-\rho-2}(m_1\alpha_n)^{\rho+1}
e^{-m_1\alpha_n}) \nonumber \\
& & =O(\alpha_n^{-1}\log^{\rho+1}{\alpha_n^{-1}}),\end{aligned}$$ where the last equality follows from (\[monen\]) and (\[expbeh\]), while in the previous one we have used the the asymptotic behavior of the incomplete gamma function $\Gamma(a,z)$ as $z\to\infty$ (see \[1; 