that $\log{y}<0$ for $0<y<1$). Further on, when computing the derivatives of (\[capitalf\]) and (\[capitalfm\]), we shall write $$F_b^{(j)}(e^{-\alpha_n}) =F_b^{(j)}(x)\mid_{x=e^{-\alpha_n}},
\quad F_{m,b}^{(j)}(e^{-\alpha_n})
=F_{m,b}^{(j)}(x)\mid_{x=e^{-\alpha_n}}, j=1,2,3.$$ Our next lemma establishes estimates on the tails $F_b^{(j)}(e^{-\alpha_n})-F_{m,b}^{(j)}(e^{-\alpha_n})$ for some specific values of $m$.

Suppose that the weight sequence $b$ satisfies conditions $(M_1)$ and $(M_2)$ and that $\alpha_n, n\ge
1$, is defined by eq. (\[alphan\]). Moreover, let $m=m(n)$ be a sequence of integers satisfying $$\label{mn}
m\sim\rho\alpha_n^{-1}\log{\alpha_n^{-1}}, \quad n\to\infty.$$ Then $$F_b^{(j)}(e^{-\alpha_n})-F_{m,b}^{(j)}(e^{-\alpha_n})
=O(\alpha_n^{-j}\log^{\rho+j}{\alpha_n^{-1}}),\quad j=1,2,3.$$

[*Proof.*]{} We shall consider only the case $j=1$. The other two cases are studied in a similar way.

First, we choose a sequence of integers $m_1(n)$ that satisfies the asymptotic equivalence $$\label{monen}
m_1=m_1(n)\sim (\rho+1)\alpha_n^{-1}\log{\alpha_n^{-1}}$$ and decompose the difference of the first derivatives in the following way: $$\label{diff}
F_b^\prime(e^{-\alpha_n})-F_{m,b}^\prime(e^{-\alpha_n})
=\sum_{k=m+1}^\infty \frac{kb_k
e^{-(k-1)\alpha_n}}{1-e^{-k\alpha_n}} =\Sigma_1+\Sigma_2,$$ where $$\Sigma_1=\sum_{k=m+1}^{m_1} \frac{kb_k
e^{-(k-1)\alpha_n}}{1-e^{-k\alpha_n}}, \quad \Sigma_2
=\sum_{k=m_1+1}^\infty \frac{kb_k
e^{-(k-1)\alpha_n}}{1-e^{-k\alpha_n}}.$$ We also notice that (\[mn\]) and (\[monen\]) imply that $$\label{expbeh}
e^{-m\alpha_n}\sim\alpha_n^\rho, \quad
e^{-m_1\alpha_n}\sim\alpha_n^{\rho+1}, \quad n\to\infty.$$ Hence, applying the result of Lemma 6, for $\Sigma_1$ we obtain $$\begin{aligned}
\label{sigmaprev}
& & \Sigma_1 =O\left(\alpha_n^\rho\sum_{k=m+1}^{m_1}kb_k\right)
\nonumber \\
& &
=O\left(\alpha_n^\rho\sum_{k=m+1}^{m_1}k^{\rho+1}(L_k-L_{k-1})\right)
+O\left(\alpha_n^\rho\sum_{k=m+1}^{m_1}k^\rho\right).\end{aligned}$$ The second summand in (\[sigmaprev\]) can be approximated by a Riemann integral using (\[mn\]) and (\[monen\]). We have $$\sum_{k=m+1}^{m_1}k^\rho
=m_1^{\rho+1}\sum_{\frac{\rho}{\rho+1}<\frac{k}{m_1}\le 1}
\left(\frac{k}{m_1}\right)^\rho\frac{1}{m_1} \sim m_1^{\rho+1}
\int_{\frac{\rho}{\rho+1}}^1 