I've been working for a long time now on how to prove a proposition given in a paper about the asymptotic normality of POT-quantile estimators. Hope somebody can help me out.
Proposition
(i) Let $X_1,\ldots,X_n$ be iid RV and $X_{1:n}\le\ldots \le X_{n:n}$ the corresponding order statistics.
(ii) Let $k=k_n$ and $p=p_n$ be sequences in $\mathbb{R}_{\ge 0}$ with $k=k_n \xrightarrow[n \to \infty]{} \infty$, $\frac{k}{n}=\frac{k_n}{n}\xrightarrow[n \to \infty]{} 0$ and $\frac{k}{np}=\frac{k_n}{np_n} \xrightarrow[n \to \infty]{} \infty$ so that $\frac{\log\left( \frac{k}{np} \right) }{\sqrt{k}} \xrightarrow[n \to \infty]{} 0.$
(iii) Let $\widehat{\gamma_k}=\widehat{\gamma_{k_n}}(X_{n-k+1:n},\ldots,X_{n:n})$ be an estimator for some $\gamma \in \mathbb{R}$. ($\widehat{\gamma_k}$ is an estimator depending on the $k$ upper order statistics.) Assume that $G_k:=\sqrt{k}(\widehat{\gamma_k}-\gamma)\xrightarrow[n \to \infty]{d} G \sim N(\mu,\sigma^2)$.
Then $$ (\ast) \quad \frac{\int_1^{\frac{k}{np}}s^{\widehat{\gamma_k}-1} \log s\, ds}{\int_1^{\frac{k}{np}}s^{\gamma-1} \log s\, ds} \xrightarrow[n \to \infty]{P} 1. $$
$\underline{\text{Hint given in the paper}}$: "The proof is direct. It follows from Taylor expansions and the fact that $\widehat{\gamma_k}=\gamma+ \frac{G_k}{\sqrt{k}}$."
My ideas so far:
Define $f(s)=s^{\frac{G_k}{\sqrt{k}}}$
Taylor expansion in $1$:
$f(s)=1+f'(\xi_s)(s-1)=1+\frac{G_k}{\sqrt{k}} \xi_s^{\frac{G_k}{\sqrt{k}}-1}(s-1)$ for some $\xi_s \in (1,s).$ Then $$(\ast) = \frac{\int_1^{\frac{k}{np}}s^{\gamma-1} s^{\frac{G_k}{\sqrt{k}}} \log s\, ds}{\int_1^{\frac{k}{np}}s^{\gamma-1} \log s\, ds} = \frac{\int_1^{\frac{k}{np}}s^{\gamma-1} (1+\frac{G_k}{\sqrt{k}} \xi_s^{\frac{G_k}{\sqrt{k}}-1}(s-1)) \log s\, ds}{\int_1^{\frac{k}{np}}s^{\gamma-1} \log s\, ds}$$
$$=1+ \frac{\int_1^{\frac{k}{np}} s^{\gamma-1} \frac{G_k}{\sqrt{k}} \xi_s^{\frac{G_k}{\sqrt{k}}-1}(s-1) \log s\, ds}{\int_1^{\frac{k}{np}}s^{\gamma-1} \log s\, ds}=:1+(\ast \ast)$$ Need to show: $(\ast \ast) \xrightarrow[n \to \infty]{P} 0.$ It is $$(\ast \ast)\le G_k \frac{log\left( \frac{k}{np} \right) }{\sqrt{k}} \frac{\int_1^{\frac{k}{np}}s^{\gamma-1} \xi_s^{\frac{G_k}{\sqrt{k}}-1}(s-1) \, ds}{\int_1^{\frac{k}{np}}s^{\gamma-1} \log s\, ds}$$ and from the Slutsky theorem I know that $G_k \frac{\log\left( \frac{k}{np} \right) }{\sqrt{k}}\xrightarrow[n \to \infty ]{d}0. $ I haven't been able to work out what happens to the rest term, though.
I know that $\frac{G_k}{\sqrt{k}}=\widehat{\gamma_k}-\gamma \xrightarrow[n\to \infty]{p} 0$ but not how fast in comparison to the interval $(1,s)$ that $\xi_k$ lies in (the upper limit of interval tends to $\infty$ with $n$ because $\frac{k}{np}$ does). Does anybody have an idea on how to solve this? The hint says "Taylor expansions", so maybe I need one more to make it work?
Thanks in advance!