First of all, we provide the proof of Theorem 3.2, for it is relatively easy.
Proof of Theorem 3.2
By Lemma 2.1, we have
$$\begin{aligned} 1-F\bigl(\alpha_{n}x^{\beta_{n}}\bigr) =&\frac{1}{\sqrt{2\pi}}\bigl( \log \bigl(\alpha_{n}x^{\beta_{n}}\bigr)\bigr)^{-1}\exp \biggl(-\frac{(\log (\alpha_{n}x^{\beta_{n}}))^{2}}{2}\biggr) \\ &{}\times\bigl(1-\bigl(\log \bigl(\alpha_{n}x^{\beta_{n}}\bigr) \bigr)^{-2}\bigr)+\mathcal{S}\bigl(\alpha_{n}x^{\beta_{n}} \bigr) \\ =:&T_{1}(x)T_{2}(x)+T_{3}(x) \end{aligned}$$
for \(x>0\), where \(T_{1}(x)=\frac{1}{\sqrt{2\pi}}(\log (\alpha_{n}x^{\beta_{n}}))^{-1}\exp(-\frac{(\log (\alpha_{n}x^{\beta_{n}}))^{2}}{2})\), \(T_{2}(x)=1-(\log (\alpha_{n}x^{\beta_{n}}))^{-2}\) and \(T_{3}(x)=\mathcal {S}(\alpha_{n}x^{\beta_{n}})\).
First, we calculate \(T_{1}(x)\). By (2.9) and (2.10), we have
$$\begin{aligned} T_{1}(x) =&\frac{1}{\sqrt{2\pi}}(\log\alpha_{n})^{-1} \exp{ \biggl(-\frac {(\log \alpha_{n})^{2}}{2} \biggr)}\bigl(1+(\log\alpha_{n})^{-1} \beta_{n}\log x\bigr)^{-1} \\ &{}\times\exp \biggl(-(\log\alpha_{n})\beta_{n}\log x- \frac{\beta^{2}_{n}\log^{2}x}{2} \biggr) \\ =&\frac{1}{nx}\bigl(1+\beta^{2}_{n}\log x \bigr)^{-1}\exp \biggl(-\frac {\beta^{2}_{n}\log^{2}x}{2} \biggr) \\ =&\frac{1}{nx}\bigl(1-\beta^{2}_{n}\log x+O\bigl( \beta^{4}_{n}\bigr)\bigr) \biggl(1-\frac{\beta^{2}_{n}\log^{2}x}{2}+O\bigl( \beta ^{4}_{n}\bigr) \biggr) \\ =&\frac{1}{nx} \biggl(1-\beta^{2}_{n} \biggl(1+\frac{1}{2}\log x\biggr)\log x+O\bigl(\beta^{4}_{n} \bigr) \biggr). \end{aligned}$$
(4.1)
Second, we estimate \(T_{2}(x)\) and \(T_{3}(x)\) for \(x>0\). By (2.10), we derive
$$\begin{aligned} T_{2}(x) &=1-\beta^{2}_{n}\bigl(1+ \beta^{2}_{n}\log x\bigr)^{-2} \\ &=1-\beta^{2}_{n}\bigl(1-2\beta^{2}_{n} \log x+O\bigl(\beta^{4}_{n}\bigr)\bigr) \\ &=1-\beta^{2}_{n}+O\bigl( \beta^{4}_{n}\bigr), \end{aligned}$$
(4.2)
and by Lemma 2.1 we have
$$\begin{aligned} T_{3}(x)&\leq\frac{3}{\sqrt{2\pi}}\bigl(\log \bigl(\alpha_{n}x^{\beta_{n}} \bigr)\bigr)^{-5}\exp{ \biggl(-\frac{(\log (\alpha_{n}x^{\beta_{n}}))^{2}}{2} \biggr)} \\ &=3\beta^{4}_{n}\bigl(1+\beta^{2}_{n} \log x\bigr)^{-4}T_{1}(x) \\ &=O\bigl(n^{-1}\beta^{4}_{n}\bigr). \end{aligned}$$
(4.3)
By (4.1)-(4.3), we have
$$ 1-F^{n}\bigl(\alpha_{n}x^{\beta_{n}}\bigr)= \frac{1}{nx} \biggl(1-\beta ^{2}_{n}\biggl(1+\biggl(1+ \frac{1}{2}\log x\biggr)\log x\biggr)+O\bigl(\beta^{4}_{n} \bigr) \biggr). $$
Thus, we obtain
$$\begin{aligned}& F^{n}\bigl(\alpha_{n}x^{\beta_{n}}\bigr)- \Phi_{1}(x) \\& \quad = \biggl(1-\frac{1}{nx}\biggl(1-\beta ^{2}_{n}\biggl(1+\biggl(1+\frac{1}{2}\log x\biggr)\log x\biggr)+O\bigl(\beta^{4}_{n}\bigr)\biggr) \biggr)^{n}-\exp\biggl(-\frac{1}{x}\biggr) \\& \quad =\exp\biggl(-\frac{1}{x}\biggr) \biggl(\exp\biggl(\frac{1}{x} \biggl(\beta^{2}_{n}\biggl(1+\biggl(1+\frac {1}{2}\log x \biggr)\log x\biggr)+O\bigl(\beta^{4}_{n}\bigr)\biggr) \biggr)-1 \biggr) \\& \quad =\exp\biggl(-\frac{1}{x}\biggr) \biggl( \beta^{2}_{n}\frac{1}{x}\biggl(1+\biggl(1+ \frac {1}{2}\log x\biggr)\log x\biggr)+O\bigl(\beta^{4}_{n} \bigr) \biggr) \end{aligned}$$
(4.4)
for large n and \(x>0\). We immediately get the result of Theorem 3.2 by (4.4). □
Proof of Theorem 3.1
By Theorem 3.2 we can prove that there exists an absolute constant \(\mathcal{C}_{1}\) such that
$$\sup_{x>0}\bigl\vert F^{n}\bigl( \alpha_{n}x^{\beta_{n}}\bigr)-\Phi_{1}(x)\bigr\vert > \frac{\mathcal {C}_{1}}{\log n}. $$
In order to obtain the upper bound for \(x>0\), we need to prove
$$\begin{aligned} (\mathrm{a})&\quad \sup_{1\leq x< \infty}\bigl\vert F^{n}\bigl(\alpha_{n}x^{\beta_{n}}\bigr)- \Phi_{1}(x)\bigr\vert < d_{1}\beta^{2}_{n}, \end{aligned}$$
(4.5)
$$\begin{aligned} (\mathrm{b})&\quad \sup_{c_{n}\leq x< 1}\bigl\vert F^{n}\bigl(\alpha_{n}x^{\beta_{n}}\bigr)- \Phi_{1}(x)\bigr\vert < d_{2}\beta^{2}_{n}, \end{aligned}$$
(4.6)
$$\begin{aligned} (\mathrm{c})&\quad \sup_{0< x< c_{n}}\bigl\vert F^{n}\bigl(\alpha_{n}x^{\beta_{n}}\bigr)- \Phi_{1}(x)\bigr\vert < d_{3}\beta^{2}_{n} \end{aligned}$$
(4.7)
for \(n>n_{0}\), where \(d_{i}>0\), \(i=1,2,3\) are absolute constants and
$$c_{n}=\frac{1}{2\log\log\alpha_{n}} $$
is positive for \(n>n_{0}\). By (2.9), we have
$$0.4(2\log n)^{1/2}< \log\alpha_{n}< (2\log n)^{1/2} $$
for \(n>n_{0}\).
First, consider the case of \(x\geq c_{n}\). Set
$$\begin{aligned}& R_{n}(x)=-\bigl[n\log F\bigl(\alpha_{n}x^{\beta_{n}} \bigr)+n\Psi_{n}(x)\bigr], \\& B_{n}(x)= \exp(-R_{n}),\qquad A_{n}(x)=\exp \biggl(-n\Psi_{n}(x)+ \frac{1}{x}\biggr), \end{aligned}$$
where \(\Psi_{n}(x)=1-F(\alpha_{n}x^{\beta_{n}})\) and \(A_{n}(x)\rightarrow1\), as \(x\rightarrow\infty\). We have
$$\begin{aligned} \Psi_{n}(x)&\leq\Psi_{n}(c_{n})< \frac{1}{\sqrt{2\pi}} \bigl(\log \bigl(\alpha_{n}c^{\beta_{n}}_{n}\bigr) \bigr)^{-1}\exp{ \biggl(-\frac{(\log (\alpha_{n}c^{\beta_{n}}_{n}))^{2}}{2} \biggr)} \\ &=\frac{1}{n}\bigl(1+\beta^{2}_{n}\log c_{n}\bigr)^{-1}\exp \biggl(-\log c_{n}- \frac{\beta^{2}_{n}\log^{2}c_{n}}{2} \biggr) \\ &< \frac{1}{n}\bigl(1+\beta^{2}_{n}\log c_{n}\bigr)^{-1}c^{-1}_{n} \\ &= \biggl(1-\frac{\log(2\log\log\alpha_{n})}{(\log\alpha _{n})^{2}} \biggr)^{-1}\frac{2\log\log\alpha_{n}}{n} \\ &< \tilde{c}_{4}< 1 \end{aligned}$$
for \(n>n_{0}\). So,
$$\inf_{x>c_{n}}\bigl(1-\Psi_{n}(x)\bigr)>1- \tilde{c}_{4}>0. $$
Since
$$-x-\frac{x^{2}}{2(1-x)}< \log(1-x)< -x $$
for \(0< x<1\), we obtain
$$\begin{aligned} 0&< R_{n}(x)\leq\frac{n\Psi^{2}_{n}(x)}{2(1-\Psi_{n}(x))}< \frac {n\Psi^{2}_{n}(c_{n})}{2(1-\Psi_{n}(x))} \\ &< \frac{n^{-1}(1+\beta^{2}_{n}\log c_{n})^{-2}c^{-2}_{n}}{2(1-\Psi_{n}(x))} \\ &< \frac{n^{-1}(1+\beta^{2}_{n}\log c_{n})^{-2}c^{-2}_{n}(\log\alpha_{n})^{2}}{2(1-\tilde{c}_{4})\beta^{-2}_{n}} \\ &=\frac{2}{\sqrt{2\pi}(1-\tilde{c}_{4})} \biggl(1-\frac{\log (2\log\log\alpha_{n})}{(\log\alpha_{n})^{2}} \biggr)^{-2} \frac{(\log \log\alpha_{n})^{2}\log\alpha_{n}}{\exp(\frac{(\log\alpha _{n})^{2}}{2})}\beta^{2}_{n} \\ &< \tilde{c}_{5}\beta^{2}_{n} \end{aligned}$$
for \(n>n_{0}\).
Hence, we have
$$n^{-1}\beta^{-2}_{n}\bigl(1+\beta^{2}_{n} \log c_{n}\bigr)^{-2}c^{-2}_{n}< \tilde{c}_{6} $$
for \(n>n_{0}\). Thus,
$$ \bigl\vert B_{n}(x)-1\bigr\vert < R_{n}< \tilde{c}_{5}\beta^{2}_{n} $$
(4.8)
for \(n>n_{0}\). By (4.8), we have
$$\begin{aligned}& \bigl\vert F^{n}\bigl(\alpha_{n}x^{\beta_{n}}\bigr)- \Phi_{1}(x)\bigr\vert \\& \quad \leq\Phi_{1}(x)B_{n}(x)\bigl\vert A_{n}(x)-1\bigr\vert +\bigl\vert B_{n}(x)-1\bigr\vert \\& \quad < \Phi_{1}(x)\bigl\vert A_{n}(x)-1\bigr\vert + \tilde{c}_{5}\beta^{2}_{n} \end{aligned}$$
(4.9)
for \(x\geq c_{n}\).
We now prove (4.5). By (2.9), (2.10), and the definition of \(A_{n}(x)\), we have
$$ A'_{n}(x)=A_{n}(x)\frac{1}{x^{2}} \biggl( \exp\biggl(-\frac{1}{2}\beta ^{2}_{n} \log^{2}x\biggr)-1 \biggr) < 0 $$
for \(x>1\). Since
$$\begin{aligned}& 0< n\gamma(\alpha_{n})< \beta^{2}_{n}\quad \mbox{and} \quad e^{x}-1\leq xe^{x}\quad \mbox{for } 0\leq x \leq1\quad \mbox{and} \\& \exp\bigl(n\gamma(\alpha_{n})\bigr)< \exp\bigl(\beta^{2}_{n} \bigr)< \exp\biggl(\frac {25}{8\log n}\biggr)< \exp\biggl(\frac{25}{8\log n_{0}}\biggr) \quad \mbox{for } n>n_{0}, \end{aligned}$$
and by (2.5), (2.9), we have
$$\begin{aligned} \sup_{x\geq1}\bigl\vert A_{n}(x)-1\bigr\vert &= \bigl\vert A_{n}(1)-1\bigr\vert \\ &=\bigl\vert \exp\bigl(n\gamma(\alpha_{n})\bigr)-1\bigr\vert \\ &\leq n\gamma(\alpha_{n})\exp\bigl(n\gamma(\alpha_{n})\bigr) \\ &\leq\tilde{c}_{7}\beta^{2}_{n} \end{aligned}$$
(4.10)
for \(n>n_{0}\).
Combining (4.9) with (4.10), we have
$$ \sup_{x\geq1}\bigl\vert F^{n}\bigl( \alpha_{n}x^{\beta_{n}}\bigr)-\Phi_{1}(x)\bigr\vert < ( \tilde {c}_{5}+\tilde{c}_{7})\beta^{2}_{n}. $$
Second, consider the situation of \(c_{n}\leq x<1\). By Lemma 2.1, we obtain
$$\begin{aligned} -n\Psi_{n}(x)+\frac{1}{x} =&-n \biggl(\frac{1}{\sqrt{2\pi }}\bigl( \log \bigl(\alpha_{n}x^{\beta_{n}}\bigr)\bigr)^{-1}\exp \biggl(-\frac{(\log (\alpha_{n}x^{\beta_{n}}))^{2}}{2}\biggr)-\gamma\bigl(\alpha_{n}x^{\beta_{n}} \bigr) \biggr)+\frac{1}{x} \\ =&-n \biggl(\frac{1}{\sqrt{2\pi}}\bigl(\log \bigl(\alpha_{n}x^{\beta_{n}} \bigr)\bigr)^{-1}\exp\biggl(-\frac{(\log (\alpha_{n}x^{\beta_{n}}))^{2}}{2}\biggr) \\ &{}-\frac{1}{\sqrt{2\pi}}\bigl(\log \bigl(\alpha_{n}x^{\beta_{n}} \bigr)\bigr)^{-3}q_{n}\bigl(\alpha_{n}x^{\beta_{n}} \bigr)\exp\biggl(-\frac {(\log (\alpha_{n}x^{\beta_{n}}))^{2}}{2}\biggr) \biggr)+\frac{1}{x} \\ =&\frac{1}{x}\bigl(1+\beta^{2}_{n}\log x \bigr)^{-1} \biggl(-\bigl(1-(\log\alpha_{n})^{-2}q_{n} \bigl(\alpha_{n}x^{\beta _{n}}\bigr) \bigl(1+\beta^{2}_{n} \log x\bigr)^{-2}\bigr) \\ &{}\times\exp\biggl(-\frac{1}{2}\beta^{2}_{n} \log^{2}x\biggr)+1+\beta^{2}_{n}\log x \biggr) \\ =&\frac{1}{x}\bigl(1+\beta^{2}_{n}\log x \bigr)^{-1}Q_{n}(x), \end{aligned}$$
where \(0< q_{n}(x)<1\) and
$$ Q_{n}(x)=- \bigl(1-\beta^{2}_{n} q_{n} \bigl(\alpha_{n}x^{\beta_{n}}\bigr) \bigl(1+\beta^{2}_{n} \log x\bigr)^{-2} \bigr)\exp\biggl(-\frac{1}{2} \beta^{2}_{n}\log^{2}x\biggr)+1+ \beta^{2}_{n}\log x. $$
Since \(e^{-x}>1-x\), as \(x>0\), we have
$$\begin{aligned} Q_{n}(x)&< -\bigl(1-\beta^{2}_{n} q_{n} \bigl(\alpha_{n}x^{\beta_{n}}\bigr) \bigl(1+\beta^{2}_{n} \log x\bigr)^{-2}\bigr) \biggl(1-\frac{1}{2}\beta^{2}_{n} \log^{2}x\biggr)+1+\beta^{2}_{n}\log x \\ &< \beta^{2}_{n}\biggl(\bigl(1+\beta^{2}_{n} \log x\bigr)^{-2}+\frac{1}{2}\log^{2}x\biggr). \end{aligned}$$
But
$$\begin{aligned} Q_{n}(x)&>\beta^{2}_{n} q_{n}\bigl( \alpha_{n}x^{\beta_{n}}\bigr) \bigl(1+\beta^{2}_{n} \log x\bigr)^{-2}+\beta^{2}_{n}\log x \\ &>\beta^{2}_{n}\log x. \end{aligned}$$
Hence, we obtain
$$\begin{aligned} \bigl\vert Q_{n}(x)\bigr\vert &< \beta^{2}_{n} \biggl(\bigl(1+\beta^{2}_{n}\log x\bigr)^{-2}+ \frac{1}{2}\log^{2}x+\vert \log x\vert \biggr) \\ &< \beta^{2}_{n} \biggl(\biggl(1-\frac{\log(2\log\log\alpha_{n})}{\log ^{2}\alpha_{n}} \biggr)^{-2}+\frac{1}{2}\log^{2}x+\vert \log x\vert \biggr) \\ &< \beta^{2}_{n}\biggl(\tilde{c}_{8}+ \frac{1}{2}\log^{2}x+\vert \log x\vert \biggr) \end{aligned}$$
for \(n>n_{0}\), where \(c_{n}\leq x<1\). Therefore,
$$\begin{aligned} \biggl\vert -n\Psi_{n}(x)+\frac{1}{x}\biggr\vert &< \beta^{2}_{n}\biggl(\tilde{c}_{8}+ \frac {1}{2}\log^{2}x+\vert \log x\vert \biggr)x^{-1} \bigl(1+\beta^{2}_{n}\log x\bigr)^{-1} \\ &< \beta^{2}_{n}\biggl(\tilde{c}_{8}+ \frac{1}{2}\log^{2}c_{n}+\vert \log c_{n} \vert \biggr)c^{-1}_{n}\bigl(1+\beta^{2}_{n} \log c_{n}\bigr)^{-1} \\ &< \tilde{c}_{9} \end{aligned}$$
for \(n\geq n_{0}\). Thus, there exists a positive number θ satisfying \(0<\theta<1\) such that
$$\begin{aligned} \Phi_{1}(x)\bigl\vert A_{n}(x)-1\bigr\vert &< \Phi_{1}(x)\exp\biggl(\theta\biggl(-n\Psi_{n}(x)+ \frac {1}{x}\biggr)\biggr)\biggl\vert -n\Psi_{n}(x)+ \frac{1}{x}\biggr\vert \\ &< \exp(\tilde{c}_{9})\beta^{2}_{n}\sup _{c_{n}\leq x< 1}\biggl\vert \biggl(\tilde{c}_{8}+ \frac{1}{2}\log^{2}x+\vert \log x\vert \biggr)x^{-1} \biggr\vert \bigl(1+\beta ^{2}_{n}\log c_{n} \bigr)^{-1} \\ &< \tilde{c}_{10}\beta^{2}_{n}. \end{aligned}$$
(4.11)
By (4.9) and (4.11), the proof of (4.6) is complete.
Third, consider the circumstance of \(0< x< c_{n}\). In this case
$$\Phi_{1}(x)< \Phi_{1}(c_{n})= \beta^{2}_{n}, $$
we have
$$\begin{aligned} \sup_{0< x< c_{n}}\bigl\vert F^{n}\bigl( \alpha_{n}x^{\beta_{n}}\bigr)-\Phi _{1}(x)\bigr\vert &< F^{n}\bigl(\alpha_{n}c^{\beta_{n}}_{n}\bigr)+ \Phi_{1}(c_{n}) \\ &< \sup_{c_{n}< x< 1}\bigl\vert F^{n}\bigl( \alpha_{n}x^{\beta_{n}}\bigr)-\Phi_{1}(x)\bigr\vert +2 \Phi _{1}(c_{n}) \\ &< (\tilde{c}_{5}+\tilde{c}_{10})\beta^{2}_{n}+ \beta^{2}_{n} \\ &< \tilde{c}_{11}\beta^{2}_{n}. \end{aligned}$$
The proof of Theorem 3.1 is finished. □