\le c(r) n^{3r} \hat{\pi}_{r}(|z|)
\]

for $z\in\R$. Here $\hat{\pi}_{r}$ being a polynomial of degree $3r$ whose coefficients depend on the coefficients of $\pi_r$ and $r$. Due to

\begin{equation*}%\label{eq.fourth}
\begin{split}
\big|\pi_r\big(\gamma_{x-h,n}(z)\big)\big|
%	=\big\pi_r\big(x(n^{-1/2}z)^{1/2},z-n^{1/2}\big)\big
\le c(r) n^{3r/2}(|x|+1)^{3r}\hat{\pi}_{r}(|z|)
\le c(r) n^{3r} \hat{\pi}_{r}(|z|),
\end{split}
\end{equation*}

the bound also holds if we replace $x$ by $x-h$.
Note that

\begin{align*}
\phi_{\Sigma}(\gamma_{x,n}(z))=\phi\big(x(n^{-1/2}z)^{1/2}\big)\,\phi_{\mu_4-1}\big(z-n^{1/2}\big).
\end{align*}

We also need to bound integrals of the form

\begin{align*}
%	\int_1^\infty \hat{\pi}_{r}(z)\phi_{\mu_4-1}\big(z-n^{1/2}\big) \dz
\int_0^\infty \hat{\pi}_{r}(z) \, z \, \phi_{\mu_4-1}\big(z-n^{1/2}\big) \dz
&\le \int_{-\infty}^\infty \hat{\pi}_{r}(z) \, |z| \, \phi_{\mu_4-1}\big(z-n^{1/2}\big) \dz\\
&\le \E\Big[\hat{\pi}_{r}(W_n) |W_n| \Big],
\end{align*}

where $W_n\sim\cN(n^{1/2},\mu_4-1)$ is a normally distributed random variable with mean $n^{1/2}$ and variance $\mu_4-1$. As the expected value is bounded up to a constant by the sum over the first $(3r+1)$ absolute moments of $W_n$,

\begin{equation}\label{int-pi-bound}
\int_0^\infty \hat{\pi}_{r}(z) \, z \, \phi_{\mu_4-1}\big(z-n^{1/2}\big) \dz
\le c(r)n^{(3r+1)/2}.
\end{equation}

Now we expand

\begin{align*}
%&=\sum_{r=0}^{{w}-2}n^{-r/2}\int_0^\infty \big|\pi_r(\gamma_{x,n}(z))-\pi_r(\gamma_{x-h,n}(z))\big| \, \phi_{\Sigma}(\gamma_{x,n}(z)) \, \big(n^{-1/2}z\big)^{1/2}\dz\\
\phi\big(x(n^{-1/2}z)^{1/2}\big)\,\phi_{\mu_4-1}\big(z-n^{1/2}\big)
\big(n^{-1/2}z\big)^{1/2}\\
&\qquad \cdot \Big| \pi_r\big(x(n^{-1/2}z)^{1/2},z-n^{1/2}\big) - \pi_r\big((x-h)(n^{-1/2}z)^{1/2},z-n^{1/2}\big) \Big|\dz.
\end{align*}

The polynomial $\pi_r\big(x(n^{-1/2}z)^{1/2},z-n^{1/2}\big)$ is a part of $\pi_r\big((x-h)(n^{-1/2}z)^{1/2},\allowbreak z-n^{1/2}\big)$ in the sense that every summand of the first polynomial is also a summand of the second polynomial. The remaining summands of $\pi_r\big((x-h)(n^{-1/2}z)^{1/2},z-n^{1/2}\big)$ all feature the factor $h(n^{-1/2}z)^{1/2}$. Therefore, $h(n^{-1/2}z)^{1/2}$ is a common factor of every summand of $\pi_r\big(x(n^{-1/2}z)^{1/2},\allowbreak z-n^{1/2}\big) - \pi_r\big((x-h)(n^{-1/2}z)^{1/2},z-n^{1/2}\big)$ and thus

\begin{align*}
%	\label{eq.pi-pi}
&\big|\pi_r\big(x(n^{-1/2}z)^{1/2},z-n^{1/2}\big) - \pi_r\big((x-h)(n^{-1/2}z)^{1/2},z-n^{1/2}\big)\big|\\
%	\label{eq.pi+1}
&\quad\le |h|(n^{-1/2}z)^{1/2} \cdot \pi_{r,2}\big((|x|\vee |x-h|\vee 1)(n^{-1/2}|z|)^{1/2},|z-n^{1/2}|\big).
\end{align*}

for a suitable polynomial $\pi_{r,2}$ with non-negative coefficients and degree at most $3r$.

Similar in spirit to <ref>, we bound

\[
\pi_{r,2}\big((|x|\vee |x-h|\vee 1)(n^{-1/2}|z|)^{1/2},|z-n^{1/2}|\big)
\le c(r) n^{3r} \hat{\pi}_{r}(z)
\]

with $\hat{\pi}_{r}$ being a polynomial of degree $3r$ whose coefficients depend on the coefficients of $\pi_r$ and $r$. Using int-pi-bound, the expression above is bounded by

\begin{align*}
&\sum_{r=0}^{{w}-2}n^{-r/2}\int_0^\infty \phi_{\mu_4-1}\big(z-n^{1/2}\big) \big(n^{-1/2}z\big)^{1/2}
\cdot |h|(n^{-1/2}z)^{1/2} c(r) n^{3r} \widehat{\pi}_{r}(z) \dz\\*
&\quad\le |h|c(w) \sum_{r=0}^{{w}-2}n^{(5r-1)/2}\int_0^\infty \phi_{\mu_4-1}\big(z-n^{1/2}\big) \, z \,
\widehat{\pi}_{r}(z) \dz\\
&\quad\le|h| c({w}) \sum_{r=0}^{{w}-2}n^{(5r-1)/2}n^{(3r+1)/2}\\
&\quad\le|h| c({w}) n^{4{w}-8}.
\end{align*}

We conclude that

\begin{equation}\label{eq.k1}
K_1 \le |h| c({w}) n^{4{w}-8}.
\end{equation}

For $K_2$, we use that $\phi$ is Lipschitz continuous with Lipschitz constant $L>0$ and get

\begin{align}\label{eq.k2}
%	\begin{split}
%&=\sum_{r=0}^{{w}-2}n^{-r/2}\int_0^\infty \big|\pi_r(\gamma_{x-h,n}(z))\big| \, \big| \phi_{\Sigma}(\gamma_{x,n}(z))-\phi_{\Sigma}(\gamma_{x-h,n}(z)) \big| \, \big(n^{-1/2}z\big)^{1/2}\dz\nonumber\\
\big|\pi_r(\gamma_{x-h,n}(z))\big|
\big(n^{-1/2}z\big)^{1/2}
\phi_{\mu_4-1}\big(z-n^{1/2}\big)\nonumber\\
&\qquad\qquad\qquad\quad \cdot \Big| \phi\big(x(n^{-1/2}z)^{1/2}\big) - \phi\big((x-h)(n^{-1/2}z)^{1/2}\big) \Big|\dz\nonumber\\
c(r) n^{3r} \hat{\pi}_{r}(z)
\big(n^{-1/2}z\big)^{1/2}
\phi_{\mu_4-1}\big(z-n^{1/2}\big)\\*
&\qquad\qquad\qquad\quad \cdot L |h|(n^{-1/2}z)^{1/2} \dz\nonumber\\
\hat{\pi}_{r}(z)
\, z \,
\phi_{\mu_4-1}\big(z-n^{1/2}\big) \dz\nonumber\\
%	&\le|h| c({w}) \sum_{r=0}^{{w}-2}n^{(5r-1)/2}n^{(3r+1)/2}\nonumber\\
&\le|h| c({w}) n^{4{w}-8}.\nonumber
%	\end{split}
\end{align}

We are left with

\begin{equation*}
%	\begin{split}
= R_n'\int_0^\infty \big(1+\|\gamma_{x,n}(z)\|\big)^{-{w}} \big(n^{-1/2}z\big)^{1/2}+ \big(1+\|\gamma_{x-h,n}(z)\|\big)^{-{w}} \big(n^{-1/2}z\big)^{1/2}\dz
%	\end{split}
\end{equation*}

where it is obviously impossible to attain a bound in $h$. Therefore, we also have to take the factor $(1+|x|)^m$ from eq.density-Hoelder-1 into account and find a bound in $n$. We only examine the first summand as the second one can be handled in a similar fashion. Thus,

\begin{align*}
%	\begin{split}
&(1+|x|)^m\int_0^\infty \big(1+\|\gamma_{x,n}(z)\|\big)^{-{w}} \big(n^{-1/2}z\big)^{1/2} \dz\\*
%&\quad\le c(m) \int_0^\infty (1+|x|^2)^{m/2} \big(1+n^{-1/2}zx^2+(z-n^{1/2})^2\big)^{-{w}/2} \big(n^{-1/2}z\big)^{1/2} \dz\\
&\quad\leq c(m) \int_0^\infty \sqrt{(1+x^2)^{m} \big(1+n^{-1/2}zx^2+(z-n^{1/2})^2\big)^{-{w}} \big(n^{-1/2}z\big) }\dz\\
&\quad= c(m) \int_0^\infty \sqrt{ \bigg( \frac{1+x^2}{1+n^{-1/2}zx^2+(z-n^{1/2})^2} \bigg)^{m}}\\*
&\quad\qquad\qquad\quad \cdot \sqrt{\big(1+n^{-1/2}zx^2+(z-n^{1/2})^2\big)^{-({w}-m)} \big(n^{-1/2}z\big) }\dz.
%	\end{split}
\end{align*}

Recalling $w\ge m$ and $|x|<2\sqrt{n}$, we split this integral into three regions that we treat separately. First,

\begin{align*}
%	\begin{split}
&\int_0^{\sqrt{n}/2} \sqrt{ \bigg( \frac{1+x^2}{1+n^{-1/2}zx^2+(z-n^{1/2})^2} \bigg)^{m}}\\*
&\qquad\quad\cdot \sqrt{\big(1+n^{-1/2}zx^2+(z-n^{1/2})^2\big)^{-({w}-m)} \big(n^{-1/2}z\big) }\dz\\*
&\quad\le \int_0^{\sqrt{n}/2} \sqrt{ \bigg( \frac{1+x^2}{(\frac{1}{2}\sqrt{n}-n^{1/2})^2} \bigg)^{m} \big((\tfrac{1}{2}\sqrt{n}-n^{1/2})^2\big)^{-({w}-m)} \cdot \tfrac12 }\dz\\
&\quad\le \int_0^{\sqrt{n}/2} \sqrt{ \bigg( \frac{1+4n}{n/4} \bigg)^{m} \big(n/4\big)^{-({w}-m)} }\dz\\
%	&\quad\le c(w)n^{-({w}-m)/2} \sqrt{n}/2 \\
&\quad\le c(w)n^{(m+1-{w})/2}
%	\end{split}%Besser gehts nicht, wie untere Schranke für x=\sqrt{n} zeigt
\end{align*}

and second by <ref>,

\begin{align*}
%	\begin{split}
&\int_{\frac{1}{2}\sqrt{n}}^{2\sqrt{n}} \sqrt{ \bigg( \frac{1+x^2}{1+n^{-1/2}zx^2+(z-n^{1/2})^2} \bigg)^{m}}\\*
&\qquad\quad\cdot \sqrt{\big(1+n^{-1/2}zx^2+(z-n^{1/2})^2\big)^{-({w}-m)} \big(n^{-1/2}z\big) }\dz\\*
&\quad\le \int_{\frac{1}{2}\sqrt{n}}^{2\sqrt{n}} \sqrt{ \bigg( \frac{1+x^2}{1+x^2/2} \bigg)^{m} \big(1+(z-n^{1/2})^2\big)^{-({w}-m)} \cdot 2 }\dz\\
&\quad\le c(m) \int_{\frac{1}{2}\sqrt{n}}^{2\sqrt{n}} \big(1+(z-n^{1/2})^2\big)^{-({w}-m)/2} \dz\\
&\quad\le \begin{cases}
c(w) \sqrt{n} , &\text{ if } {w}=m,\\
c(w) \log n , &\text{ if } {w}=m+1.
%c(w) , &\text{ if } m \le {w}-2,
\end{cases}
%	\end{split}
\end{align*}

\begin{align*}
%	\begin{split}
&\int_{2\sqrt{n}}^\infty \sqrt{ \bigg( \frac{1+x^2}{1+n^{-1/2}zx^2+(z-n^{1/2})^2} \bigg)^{m}}\\* &\qquad\quad\cdot\sqrt{\big(1+n^{-1/2}zx^2+(z-n^{1/2})^2\big)^{-({w}-m)} \big(n^{-1/2}z\big) }\dz\\*
&\quad\le \int_{2\sqrt{n}}^\infty \sqrt{ (1+4n)^{m} \big((z-n^{1/2})^2\big)^{-{w}} \big(n^{-1/2}z\big) }\dz\\
&\quad\le c(m) n^{m/2} n^{-1/4} \int_{2\sqrt{n}}^\infty \sqrt{ \big((z-z/2)^2\big)^{-{w}} \cdot z}\dz\\
&\quad\le c(m) n^{m/2} n^{-1/4} \int_{2\sqrt{n}}^\infty z^{-(2{w}-1)/2} \dz\\
%	&\quad\le c(m) n^{m/2} n^{-1/4} n^{-(2{w}-3)/4} \dz\\
&\quad\le c(w) n^{(m+1-{w})/2}.
%	\end{split}
\end{align*}

This also holds if $x$ is replaced by $x-h$ and thus combining the last four equations and eq.def-Rn'

\begin{align}\label{eq.k3}
%	\begin{split}
(1+|x|)^m K_3
&= (1+|x|)^m R_n' \int_0^\infty \big(1+\|\gamma_{x,n}(z)\|\big)^{-{w}} \big(n^{-1/2}z\big)^{1/2}\nonumber\\*
&\qquad\qquad\qquad\qquad+ \big(1+\|\gamma_{x-h,n}(z)\|\big)^{-{w}} \big(n^{-1/2}z\big)^{1/2}\dz\nonumber\\
&\le R_n' \begin{cases}
c(w) \sqrt{n} , &\text{ if } {w}=m,\\
c(w) \log n , &\text{ if } {w}=m+1,
%c(w) , &\text{ if } m \le {w}-2,
\end{cases}\\&
\le \begin{cases}
\co\big(n^{-(m-3)/2}\big), &\text{ if } {w}=m,\nonumber\\
\co\big(n^{-(m-1)/2} \, \log n \big), &\text{ if } {w}=m+1,
%\co\big(n^{-({w}-2)/2}\big), &\text{ if } m \le {w}-2,
\end{cases}\nonumber\\*
%	\end{split}
\end{align}

uniformly for all $|x|<2\sqrt{n}$, $|h|<1$. In summary, we get by eq.f_n-f_n-Z,eq.k1,eq.k2,eq.k3,

\begin{equation*}
%	\begin{split}
\le c(w) (1+|x|)^{m} |h| n^{4{w}-8} + r_n.
%	\end{split}
\end{equation*}

Hence, setting $a=1$ and $b=4{w}-8$, the Hölder-type condition $(iii)$ in <ref> is satisfied.
It remains to derive a uniform bound on $f_n$. For this aim we subsequently use the identity for the normal density

\begin{align*}
\phi_{\sigma^2}(x)=\phi_{\sigma^2/2}(x)\,\phi_{\sigma^2/2}(x) \sqrt{2/(\pi\sigma^2)}
\end{align*}

for $x\in\R$ and $\sigma^2>0$.
By eq.k3,eq.f_n-von-g_n, it suffices to bound

\begin{align*}
\lefteqn{\int_0^\infty \sum_{r=0}^{{w}-2}n^{-r/2}\pi_r(\gamma_{x,n}(z))\phi_{\Sigma}(\gamma_{x,n}(z)) \big(n^{-1/2}z\big)^{1/2} \dz}\quad\\*
%	&=\int_0^\infty \sum_{r=0}^{{w}-2}n^{-r/2}\pi_r(\gamma_{x,n}(z)) \phi\big(x(n^{-1/2}z)^{1/2}\big) \, \phi_{\mu_4-1}\big(z-n^{1/2}\big) \big(n^{-1/2}z\big)^{1/2} \dz\\
&= \sqrt{2/(\pi(\mu_4-1))} \int_0^\infty \sum_{r=0}^{{w}-2}n^{-r/2}\pi_r(\gamma_{x,n}(z)) \, \phi\big(x(n^{-1/2}z)^{1/2}\big) \, \phi_{(\mu_4-1)/2}\big(z-n^{1/2}\big)  \\*
&\qquad \qquad \qquad \qquad \qquad \qquad \cdot \phi_{(\mu_4-1)/2}\big(z-n^{1/2}\big) \, \big(n^{-1/2}z\big)^{1/2} \dz\\
&= \sqrt{2/(\pi(\mu_4-1))} \int_0^\infty \sum_{r=0}^{{w}-2}n^{-r/2}\pi_r(\gamma_{x,n}(z)) \phi_{\Sigma_2}(\gamma_{x,n}(z)) \\*
&\qquad \qquad \qquad \qquad \qquad \qquad \cdot \phi_{(\mu_4-1)/2}\big(z-n^{1/2}\big) \, \big(n^{-1/2}z\big)^{1/2} \dz\\
%	&\le n^{-1/4}\int_0^\infty \sum_{r=0}^{{w}-2}n^{-r/2} c(r) \, \phi_{(\mu_4-1)/2}\big(z-n^{1/2}\big) \, z^{1/2} \dz\\
&\le c(w) n^{-1/4}\int_0^\infty \phi_{(\mu_4-1)/2}\big(z-n^{1/2}\big) \, z^{1/2} \dz
\end{align*}

\begin{align*}
\Sigma_2=
\begin{pmatrix}
1\enskip & 0\\
0\enskip & (\mu_4-1)/2
\end{pmatrix}.
\end{align*}

The integral is divided into two parts that are evaluated separately. First,

\begin{align*}
c(w) n^{-1/4}\int_0^{2n^{1/2}} \phi_{(\mu_4-1)/2}\big(z-n^{1/2}\big) \, z^{1/2} \dz
&\le c(w) \int_0^{2n^{1/2}} \phi_{(\mu_4-1)/2}\big(z-n^{1/2}\big)  \dz\\
&\le c(w)
\end{align*}

and second,

\begin{align*}
\lefteqn{c(w) n^{-1/4}\int_{2n^{1/2}}^\infty \phi_{(\mu_4-1)/2}\big(z-n^{1/2}\big) \, z^{1/2} \dz}\quad\\
&= c(w) n^{-1/4}\int_{n^{1/2}}^\infty \phi_{(\mu_4-1)/2}(z) \, \big(z+n^{1/2}\big)^{1/2} \dz\\
%	&\le  c(w) n^{-1/4}\int_{n^{1/2}}^\infty \phi_{(\mu_4-1)/2}(z) \, z^{1/2} \dz\\
&\le  c(w) n^{-1/4}\int_{n^{1/2}}^\infty \phi_{(\mu_4-1)/2}(z) \, z^{2} \dz \le  c(w).
\end{align*}

Thus, $f_n$ is bounded uniformly over all $n\ge N_1$ and <ref> yields

\begin{equation*}
\sup_{x\in\R} \, (1+|x|)^{m}|f_n(x)-\phi^{q}_{m,n}(x)|
= \co\big( n^{-({s}-2)/2} \, (\log n)^{({s}+2\lfloor {s} \rfloor)/2} \big)
+ \cO\big( n^{- \lceil \frac{m-1}2 \rceil} \big)
+ r_n
\end{equation*}

for both $m =w-1$ and $m=w$, and for all ${s}$ with $m \le {s} \le 2{w}$. We choose ${s}=m+2$ to finally get

\begin{align*}
\sup_{x\in\R} \, (1+|x|)^{m}|f_n(x)-\phi^{q}_{m,n}(x)|
%	&= \begin{cases}
%	\cO\big( n^{- \lceil \frac{m-1}2 \rceil} \big) + \co\big(n^{-({w}-3)/2}\big), &\text{ if } {w}=m,\\
%	\cO\big( n^{- \lceil \frac{m-1}2 \rceil} \big) + \co\big(n^{-({w}-2)/2} \, \log n \big), &\text{ if } {w}=m+1,
%\cO\big( n^{- \lceil \frac{m-1}2 \rceil} \big) + \co\big(n^{-({w}-2)/2}\big), &\text{ if } m \le {w}-2.
%	\end{cases}\\
%	&= \begin{cases}
%	\co\big(n^{-({w}-3)/2}\big), &\text{ if } {w}=m,\\
%	\co\big(n^{-({w}-2)/2} \, \log n \big), &\text{ if } {w}=m+1,
%	%\cO\big( n^{- \lceil \frac{m-1}2 \rceil} \big) , &\text{ if } m \le {w}-2,
%	\end{cases}\\
&= \begin{cases}
\co\big(n^{-(m-3)/2}\big), &\text{ if } {w}=m,\\
\co\big(n^{-(m-1)/2} \, \log n \big), &\text{ if } {w}=m+1.
%\cO\big( n^{- \lceil \frac{m-1}2 \rceil} \big) , &\text{ if } m \le {w}-2
\end{cases}
\end{align*}

§ RATE OF CONVERGENCE AND EDGEWORTH-TYPE EXPANSION IN THE ENTROPIC CENTRAL LIMIT THEOREM FOR SELF-NORMALIZED SUMS

The main result of this section is the following theorem.

Assume that $X_1$ is symmetric, the distribution of $X_1$ is non-singular, <ref> is satisfied and $\E|X_1|^{2m}<\infty$ for some $m\in\N$, $m\ge3$. Then

\begin{equation}\label{eq.t-entropy-general}
D(T_n)=\frac{c_2}{n^2}+\dots+\frac{c_{\lfloor (m-2)/2 \rfloor}}{n^{\lfloor (m-2)/2 \rfloor}}+ \co\big((n\log n)^{-(m-2)/2}\,\log n\big),
\end{equation}

\begin{align*}%\label{eq.c_}
c_l=\sum_{k=2}^{2l} \frac{(-1)^k}{k(k-1)} \sum \int_{\R} \frac{q_{r_1}(x)\dots q_{r_k}(x)}{\phi(x)^{k-1}} \dx, \quad l\in\N,
\end{align*}

with the inner sum running over all positive integers $r_1,\dots,r_k$ such that $r_1+\dots+r_k=2l$.

Together with <ref>, <ref> implies <ref>. Before starting with the proof of <ref>, let us record the following observation.

The classical statistic $Z_n$ is normalized in the sense that $\E Z_n=0$ and $\Var Z_n=1$. The $t$-statistic is not normalized (see <cit.>). In general, the self-normalized sum is also not normalized (see <ref>). If $X_1$ is symmetric however, it is normalized because

\begin{equation*}
\E T_n
=\E \Big[\tE \big[ S_n/V_n \big]\Big]
=\E \Big[V_n^{-1}\sum \tE \big[ X_j \big] \Big]
\end{equation*}

\begin{equation*}
\E T_n^2
=\E \Big[\tE \big[ S_n^2/V_n^2 \big]\Big]
=\E \Big[V_n^{-2}\Big(\sum \tE \big[ X_j^2 \big] + 2\sum_{j=1}^{n-1}\sum_{k=j+1}^{n} \tE \big[ X_jX_k \big]\Big) \Big]
\end{equation*}

In the following proof, we adapt some ideas from [13].

By <ref>, the densities $f_n$ exist and are uniformly bounded by some constant $M$ for $n\ge N$. Throughout the remainder of the proof assume $n\ge N$.
Since $X_1$ is symmetric, $T_n$ is normalized (see <ref>). Therefore, its relative entropy, defined in eq.def-rel-entropy, includes the standard normal density $\phi_{0,1}=\phi$.

Set $\Delta_{m,n}=(n\log n)^{-(m-2)/2}\,\log n$.
Let $A_n\ge1$, then we split up the integral from the definition

\begin{align}\label{eq.D-ints}
= \int_{|x|\le A_n} f_n(x)\log \frac{f_n(x)}{\phi(x)}\dx
+ \int_{|x|>A_n} f_n(x)\log \frac{f_n(x)}{\phi(x)}\dx.
\end{align}

For the second integral, we get the upper bound

\begin{align*}
\int_{|x|>A_n} f_n(x)\log \frac{f_n(x)}{\phi(x)}\dx
&\le\int_{|x|>A_n} f_n(x)\log \frac{M}{\phi(x)}\dx\\
%&=\int_{|x|>A_n} f_n(x)\big(\log \big(M\sqrt{2\pi}\big)+x^2/2\big)\dx\\
&=\log \big(M\sqrt{2\pi}\big)\int_{|x|>A_n} f_n(x)\dx+\tfrac12 \int_{|x|>A_n} x^2 f_n(x)\dx\\
%&\le \log \big(M\sqrt{2\pi}\big)A_n^{-2}\int_{|x|>A_n} x^2 f_n(x)\dx+\tfrac12 \int_{|x|>A_n} x^2 f_n(x)\dx\\
&\leq\Big(\log \big(M\sqrt{2\pi}\big)A_n^{-2}+\tfrac12\Big) \int_{|x|>A_n} x^2 f_n(x)\dx.
\end{align*}

By means of the inequality $u\log u\ge u-1$ for $u>0$, we get the lower bound

\begin{align*}
\int_{|x|>A_n} f_n(x)\log \frac{f_n(x)}{\phi(x)}\dx
&=\int_{|x|>A_n} \phi(x) \frac{f_n(x)}{\phi(x)} \log \frac{f_n(x)}{\phi(x)}\dx\\
&\ge\int_{|x|>A_n} \phi(x) \Big(\frac{f_n(x)}{\phi(x)} -1\Big)\dx\\
&=\int_{|x|>A_n} \big( f_n(x)-\phi(x) \big) \dx\\
%&=\P\{|T_n|>A_n\} - \P\{|Z|>A_n\}\\
&\ge - \P\{|Z|>A_n\}.
\end{align*}

\begin{align}\label{eq.entropy>A_n}
\bigg|\int_{|x|>A_n} f_n(x)\log \frac{f_n(x)}{\phi(x)}\dx\bigg|
\le C \int_{|x|>A_n} x^2 f_n(x)\dx
\end{align}

where $C=\log \big(M\sqrt{2\pi}\big)+1/2$. Before further evaluation of this term, we specify $A_n$ with the intent to balance both integrals in eq.D-ints.

\begin{align*}
A_n=\sqrt{(m-2)\log n + (m-3)\log\log n + \rho_n},
\end{align*}

where $\rho_n\to \infty$ sufficiently slow and $\rho_n\le\log n$. $(\rho_n)$ is needed to achieve the order $\co$ and not only $\cO$ in the remainder in eq.t-entropy-general. It will be defined at a later stage, depending on the remainder in <ref>. For any $k\ge0$, the bound

\begin{equation}\label{eq.A_n-exp}
\begin{split}
A_n^k \, e^{-A_n^{2}/2}
&= \big((m-2)\log n + (m-3)\log\log n + \rho_n\big)^{k/2}\\*
&\quad \cdot n^{-(m-2)/2} (\log n)^{-(m-3)/2} \, e^{-\rho_n/2}\\
&=\co\big(n^{-(m-2)/2} (\log n)^{-(m-3-k)/2}\big)
\end{split}
\end{equation}

is frequently needed.

Recall that $|T_n|\le\sqrt{n}$ by <ref>. We return to eq.entropy>A_n and derive by partial integration for Stieltjes integrals (see e.g. <cit.>) using the symmetry of $T_n$,

\begin{align}\label{eq.entropy>A_n-expanded}
%	\begin{split}
\int_{|x|>A_n} x^2 f_n(x)\dx
%	&= 2\int_{A_n}^{\sqrt{n}} x^2 f_n(x)\dx\nonumber\\
&= 2\int_{A_n}^{\sqrt{n}} x^2 \dd F_n(x)\nonumber\\
&= 2\Big(\sqrt{n}^2 (F_n(\sqrt{n} \, ))-A_n^2 (F_n(A_n))-\int_{A_n}^{\sqrt{n}} F_n(x) 2x \dx \Big)\\
%	&= 2n-2A_n^2\big(F_n(A_n)\big)-2\int_{A_n}^{\sqrt{n}} 2x \big( F_n(x)\big)\dx\nonumber\\
%&= 2A_n^2 \big(1-F_n(A_n)\big)+2\int_{A_n}^{\sqrt{n}} 2x \big( 1-F_n(x)\big)\dx\\
&= 2A_n^2 \big(1-F_n(A_n)\big)+4\int_{A_n}^{\infty} x \big( 1-F_n(x)\big)\dx\nonumber %ob \sqrt{n} oder \infty ist hier egal
%	\end{split}
\end{align}

By <ref> (with $s=m+1$),

\begin{equation*}
F_n(x) = \Phi^{Q}_{m+1,n}(x) + (1+|x|)^{-(m+1)} R_n(x)
\end{equation*}

for all $x\in\R$ and $n$ sufficiently large. Here, $\sup_{x} |R_n(x)|=\co\big( n^{-(m-1)/2} \, (\log n)^{(3m+3)/2} \big)$.

Therefore, we can replace $A_n^2\big(1-F_n(A_n)\big)$ with

\begin{equation*}
\end{equation*}

with an error of magnitude

\begin{align*}
=\co\big( n^{-(m-1)/2} \, (\log n)^{(2m+4)/2} \big).
\end{align*}

Recall the definition

\begin{align*}
\Phi^{Q}_{m+1,n}(x)=\Phi(x)+\sum_{r=1}^{m-1}Q_{r}(x)n^{-r/2}
\end{align*}

from eq.Phi^Q-def. For uneven $r$, $Q_r$ vanishes and for even $r$, $Q_{r}$ has the useful form of $\phi$ multiplied with a polynomial of degree $2r-1$. The coefficients of $Q_{r}$ are functions of the moments $\mu_3,\dots,\mu_{r+2}$ and can thus be bounded by $c(r)$. Now by eq.A_n-exp,

\begin{align*}
\Big|A_n^2\sum_{r=1}^{m-1}Q_{r}(A_n)n^{-r/2}\Big|
\le A_n^2 \, c(m) \, A_n^{2m-3} \phi(A_n) n^{-1/2}
=\co\big(n^{-(m-1)/2} (\log n)^{(m+2)/2}\big).
%\le c(m) \phi(A_n)
\end{align*}

For the last part, we use $1-\Phi(x)\le \phi(x)/x$ (for $x>0$) and eq.A_n-exp to achieve

\begin{align*}
%	=2 A_n^2 \big(1-\Phi(A_n)\big)
%	\le  A_n^2 \frac{\phi(A_n)}{A_n}
\le A_n \phi(A_n)
%=\co\big(n^{-(m-2)/2} (\log n)^{-(m-4)/2}\big)
\end{align*}

and therefore

\begin{align*}
%=\co\big(n^{-(m-2)/2} (\log n)^{-(m-4)/2}\big).
\end{align*}

Similarly, in the second part of eq.entropy>A_n-expanded, we can replace $\int_{A_n}^{\infty} x \big( 1-F_n(x)\big)\dx$ with

\begin{align*}
\int_{A_n}^{\infty} x \big( 1-\Phi^{Q}_{m+1,n}(x)\big)\dx
\end{align*}

with an error of at most

\begin{align*}
\int_{A_n}^{\infty} x (1+|x|)^{-(m+1)} R_n(x)\dx
=\co\big( n^{-(m-1)/2} \, (\log n)^{(2m+4)/2} \big).
\end{align*}

As above, using $\alpha=\frac{1}{2(2m-2)}$, <ref> (with $\beta=1$, $\nu_n=n^\alpha$) and eq.A_n-exp,

\begin{align}\label{eq.int-Q}
%	\begin{split}
\lefteqn{\Big|\int_{A_n}^{\infty} x\sum_{r=1}^{m-1}Q_{r}(x)n^{-r/2} \dx\Big|}\quad\nonumber\\*
%&\le c(m) n^{-1/2} \int_{A_n}^{\infty} x^{2m-2} \phi(x) \dx\nonumber\\
&\leq c(m) n^{-1/2} \Big(\int_{A_n}^{n^\alpha} x^{2m-2} \phi(x) \dx + \int_{n^\alpha}^{\infty} x^{2m-2} \phi(x) \dx \Big)\nonumber\\
&\le c(m) \int_{A_n}^{n^\alpha} \phi(x) \dx + c(m) \exp\big(-\tfrac14 n^{2\alpha}\big)\\
%&\le c(m) \big(1-\Phi(A_n)\big) + c(m) \exp\big(-\tfrac14 n^{2\alpha}\big)\nonumber\\
&\le c(m) \phi(A_n) + c(m) \exp\big(-\tfrac14 n^{2\alpha}\big)\nonumber\\
&=\co\big(n^{-(m-2)/2} (\log n)^{-(m-3)/2}\big)\nonumber
%	\end{split}
\end{align}

and by eq.A_n-exp,

\begin{align*}
%	\int_{A_n}^{\infty} x \big( \Phi(-x)+1-\Phi(x)\big)\dx
\int_{A_n}^{\infty} x \big( 1-\Phi(x)\big)\dx
&\le  \int_{A_n}^{\infty} \phi(x) \dx\le \phi(A_n)
=\co\big(n^{-(m-2)/2} (\log n)^{-(m-3)/2}\big)
\end{align*}

and therefore

\begin{align*}
\int_{A_n}^{\infty} x \big( 1-F_n(x)\big)\dx
=\co\big(n^{-(m-2)/2} (\log n)^{-(m-3)/2}\big).
\end{align*}

Hence by eq.entropy>A_n-expanded,

\begin{align}\label{eq.int-x^2f_n}
\int_{|x|>A_n} x^2 f_n(x)\dx
%=\co\big(n^{-(m-2)/2} (\log n)^{-(m-4)/2}\big)
\end{align}

and by inserting

\begin{align}\label{eq.P-Z>A_n}
\P\{|Z|>A_n\}
%= \Phi(-A_n)+1-\Phi(A_n)
\le 2 \phi(A_n)
=\co\big(n^{-(m-2)/2} (\log n)^{-(m-3)/2}\big)
\end{align}

into eq.entropy>A_n, we obtain

\begin{align*}
\bigg|\int_{|x|>A_n} f_n(x)\log \frac{f_n(x)}{\phi(x)}\dx\bigg|
%=\co\big(n^{-(m-2)/2} (\log n)^{-(m-4)/2}\big).
\end{align*}

In order to evaluate $D(T_n)$, we thus only have to examine (with $L(u):=u\log(u)$)

\begin{equation}\label{eq.entropy<A_n}
\begin{split}
\int_{|x|\le A_n} f_n(x)\log \frac{f_n(x)}{\phi(x)}\dx
&= \int_{|x|\le A_n} L\Big(\frac{f_n(x)}{\phi(x)}\Big)\phi(x) \dx\\
&= \int_{|x|\le A_n} L\big(1+u_m(x)+v_n(x)\big)\phi(x) \dx,
\end{split}
\end{equation}

\begin{align*}
\qquad \text{and}\qquad
\end{align*}

By <ref>,

\begin{equation}\label{eq.entropy-llt}
\begin{split}
\big| f_n(x) - \phi^{q}_{m-1,n}(x)\big|
&\le (1+|x|)^{-(m-1)} R_n(x) \\
&\le (1+|x|)^{-(m-1)} n^{-(m-2)/2} \log(n) r_n
\end{split}
\end{equation}

for all $x\in\R$ and $n$ sufficiently large. Here, $r_n=n^{(m-2)/2} (\log n)^{-1} \, \sup_{x} |R_n(x)| =\co(1)$.

There exists $C_m>0$, such that $y\mapsto (1+y)^{m-1}\phi(y)$ is decreasing for $y\ge C_m$ and thus its inverse is increasing. Additionally, for $|x|\le C_m$, we bound

\[
(1+|x|)^{-(m-1)}\phi(x)^{-1}\le c(m).
\]

Therefore, we estimate

\[
(1+|x|)^{-(m-1)}\phi(x)^{-1}\le (1+A_n)^{-(m-1)}\phi(A_n)^{-1}
\]

for all $|x|\le A_n$ and for $n$ sufficiently large. Hence,

\begin{align}\label{eq.vn-bound}
&\le (1+|x|)^{-(m-1)} \phi(x)^{-1} n^{-(m-2)/2} \log(n) r_n\nonumber\\
&\le c(m) A_n^{-(m-1)} e^{A_n^2/2} \, n^{-(m-2)/2} \log(n) r_n\nonumber\\
&= c(m) \big((m-2)\log n + (m-3)\log\log n + \rho_n\big)^{-(m-1)/2} \big(\log n\big)^{(m-3)/2} \\
&\quad \cdot e^{\rho_n/2} \log(n) r_n\nonumber\\
&\le c(m) e^{\rho_n/2} r_n.\nonumber
\end{align}

Next, we choose the sequence $(\rho_n)$ in a way that the last expression goes to 0. Thus, ${|v_n(x)|<1/4}$ holds for $|x|\le A_n$ and $n$ sufficiently large.

Recall $\phi^{q}_{m-1,n}(x)=\phi(x)+\sum_{r=1}^{m-3}q_{r}(x)n^{-r/2}$ from eq.phi^q-def. For even $r$, $q_{r}$ factorizes as $\phi$ multiplied with an even polynomial of degree $2r$. The coefficients of $q_{r}$ are functions of the moments $\mu_3,\dots,\mu_{r+2}$ and can thus be bounded by $c(r)$. As above, $q_{r}=0$ for uneven $r\in\N$. So

\begin{equation}\label{eq.u_m}
\begin{split}
\le c(m)\big(1+|x|^{2m-6}\big)n^{-1/2}
\le c(m)A_n^{2m-6} n^{-1/2},
\end{split}
\end{equation}

where the last inequality only holds for $|x|\le A_n$. In particular $|u_m(x)|<1/4$ holds for $|x|\le A_n$ and $n$ sufficiently large.
Next, by Lemma <ref>,

\begin{equation*}%\label{eq.Taylor-L}
L(1+u+v)=L(1+u)+v+\theta_1 u v+\theta_2 v^2
\end{equation*}

for $|u|\le 1/4, |v|\le1/4$ and $|\theta_j|\le2$ depending on $u$ and $v$.
Inserting $u=u_m(x)$ and $v=v_n(x)$, shows how to eliminate $v_n(x)$ from eq.entropy<A_n. This produces an error not exceeding

\[
\]

\begin{align*}
&=\int_{|x|\le A_n}\big(f_n(x)-\phi^{q}_{m-1,n}(x)\big)\dx,\\
&=\int_{|x|\le A_n}\big|u_m(x)\big|\big|f_n(x)-\phi^{q}_{m-1,n}(x)\big|\dx \\
\intertext{and}
&=\int_{|x|\le A_n}\frac{\big(f_n(x)-\phi^{q}_{m-1,n}(x)\big)^2}{\phi(x)}\dx.
\end{align*}

For even $r\in\N$, the function $q_{r}$ factorizes as $\phi$ multiplied with a sum of Hermite polynomials $H_k$ for $k>0$ (see eq.tp,eq.E-tp-q,eq.q-2). Now

\begin{align*}
\int_\R \phi(x)H_k(x)\dx=\int_\R \phi(x)H_k(x)H_0(x)\dx=0
\end{align*}

for all $k>0$ yields

\begin{align}\label{eq.int-q=0}
\int_\R q_{r}(x)\dx=0
\end{align}

for all even $r=1,\dots,m-3$. As $q_{r}=0$ for uneven $r\in\N$,

\begin{align*}
\int_\R\ \big(f_n(x)-\phi^{q}_{m-1,n}(x)\big)\dx=0.
\end{align*}

\begin{equation*}
\le \int_{|x|>A_n}f_n(x)\dx + \int_{|x|>A_n}\big|\phi^{q}_{m-1,n}(x)\big|\dx,
\end{equation*}

where the first integral is of order $\co(\Delta_{m,n})$ by eq.int-x^2f_n. Using eq.u_m and eq.P-Z>A_n, the second integral is bounded by

\begin{equation}\label{eq.int1}
\begin{split}
&\int_{|x|>A_n}\big|\phi^{q}_{m-1,n}(x)-\phi(x)\big|\dx +\int_{|x|>A_n}\phi(x)\dx\\
&\quad \le c(m)\int_{|x|>A_n}|x|^{2m-6} n^{-1/2} \phi(x)\dx +\P\{|Z|>A_n\}\\
&\quad =\co\big(n^{-(m-2)/2} (\log n)^{-(m-3)/2}\big).
\end{split}
\end{equation}

For $m>3$, we have used eq.int-Q with $\alpha=\frac{1}{2(2m-6)}$ and for $m=3$ the order is true because in that case the order of the first term is smaller than the second one.
Therefore, ${|D_1|=\co(\Delta_{m,n})}$. By eq.u_m and eq.entropy-llt,

\begin{align*}
%&=\int_{|x|\le A_n}\big|u_m(x)\big|\big|f_n(x)-\phi^{q}_{m-1,n}(x)\big|\dx\\
&\le c(m) \int_{|x|\le A_n} A_n^{2m-6} n^{-1/2} (1+|x|)^{-(m-1)} n^{-(m-2)/2} \log(n) r_n \dx\\
&\le c(m) A_n^{2m-5} n^{-(m-1)/2} \log(n) r_n\\
&=\co(n^{-(m-1)/2} (\log n)^{(2m-3)/2}).
\end{align*}

Similarly as in eq.vn-bound, we obtain for all $|x|\le A_n$

\begin{align*}
%&=\int_{|x|\le A_n}\frac{\big(f_n(x)-\phi^{q}_{m-1,n}(x)\big)^2}{\phi(x)}\dx\\
&\le \int_{|x|\le A_n} (1+|x|)^{-2(m-1)} \phi(x)^{-1} n^{-(m-2)} (\log n)^2 r_n^2 \dx\\
&\le c(m) A_n^{-(2m-3)} \phi(A_n)^{-1} n^{-(m-2)} (\log n)^2 r_n^2 \\
&= c(m) \big((m-2)\log n + (m-3)\log\log n + \rho_n\big)^{-(2m-3)/2} \big(\log n\big)^{(m-3)/2} \, e^{\rho_n/2} \\
&\quad \cdot n^{-(m-2)/2} (\log n)^2 r_n^2 \\
&\le c(m) (\log n)^{-(m-4)/2} n^{-(m-2)/2} r_n\\
%&=\co\big(n^{-(m-2)/2} (\log n)^{-(m-4)/2}\big).
\end{align*}

Thus, eliminating $v_n(x)$ from eq.entropy<A_n leads to

\begin{equation*}
D(T_n)= \int_{|x|\le A_n} L\big(1+u_m(x)\big)\phi(x) \dx + \co(\Delta_{m,n}).
\end{equation*}

For $m=3$, $u_m(x)=0$ and eq.t-entropy-general holds. Hence, assume $m\ge4$ from now on. By Taylor expansion around $u=0$,

\begin{align*}
%&= (1+0)\log(1+0) + u \big(\log(1+0) + 1\big) + \tfrac{u^2}{2!} (1+0)^{-1} + \tfrac{u^3}{3!} (-1)^1 1! (1+0)^{-2}\\
%&\quad + \tfrac{u^4}{4!} (-1)^2 2! (1+0)^{-3} + \dots + \tfrac{u^{m-2}}{(m-2)!} (-1)^{m-4} (m-4)! (1+0)^{-(m-3)}\\
%&\quad + \tfrac{u^{m-1}}{(m-1)!} (-1)^{m-3} (m-3)! (1+\xi u)^{-(m-2)}\\
&= u + \sum_{k=2}^{m-2} \frac{(-1)^k}{k(k-1)}u^k+\theta u^{m-1}
\end{align*}

for some $\theta \ge 0$, depending on $u$ and $m$ that can be bounded by $c(m)$ for all $|u|\le1/4$.

Inserting $u_m$ yields

\begin{align*}
\lefteqn{\int_{|x|\le A_n} L\big(1+u_m(x)\big)\phi(x) \dx}\quad\\
&\le\int_{|x|\le A_n} \phi^{q}_{m-1,n}(x)-\phi(x) \dx
+ \sum_{k=2}^{m-2} \frac{(-1)^k}{k(k-1)} \int_{|x|\le A_n} u_m(x)^k\phi(x) \dx \\
&\quad+ c(m) \int_{|x|\le A_n} \big|u_m(x)\big|^{m-1} \phi(x) \dx\\
\end{align*}

for $n$ large enough. By eq.int-q=0 and eq.int1,

\begin{align*}
\lefteqn{|E_1|=\bigg| \int_{|x|\le A_n} \phi^{q}_{m-1,n}(x)-\phi(x) \dx \bigg|
= \bigg| \int_{|x|> A_n} \phi^{q}_{m-1,n}(x)-\phi(x) \dx \bigg|}\quad\\*
&\le \int_{|x|> A_n} \big|\phi^{q}_{m-1,n}(x)-\phi(x)\big| \dx
=\co\big(n^{-(m-2)/2} (\log n)^{-(m-3)/2}\big)
\end{align*}

and by eq.u_m

\begin{align*}
%&\int_{|x|\le A_n} \big|u_m(x)\big|^{m-1} \phi(x) \dx\\
&\le \int_\R \big|u_m(x)\big|^{m-1} \phi(x) \dx\\
&\le c(m) n^{-(m-1)/2}\int_\R \big(1+|x|^{2m-6}\big)^{m-1} \phi(x) \dx=\cO\big(n^{-(m-1)/2}\big).
\end{align*}

By eq.u_m and eq.int-Q with $\alpha=\frac{1}{2(2m-6)}$, the integral in $E_2$ can be extended to the whole real line at the expense of an error not exceeding

\begin{align*}
\lefteqn{\sum_{k=2}^{m-2} \frac{(-1)^k}{k(k-1)} \int_{|x|> A_n} u_m(x)^k\phi(x) \dx}\quad\\
&\le c(m) \sum_{k=2}^{m-2} \frac{(-1)^k}{k(k-1)} n^{-k/2} \int_{|x|> A_n} \big(1+|x|^{2m-6}\big)^k \phi(x) \dx\\
&=\co\big(n^{-(m-1)/2} (\log n)^{-(m-3)/2}\big).
\end{align*}

In summary,

\begin{align*}
&= \int_{|x|\le A_n} L\big(1+u_m(x)\big)\phi(x) \dx + \co(\Delta_{m,n})\\*
&= \sum_{k=2}^{m-2} \frac{(-1)^k}{k(k-1)} \int_{\R} \frac{\big(\phi^{q}_{m-1,n}(x)-\phi(x)\big)^k}{\phi(x)^{k-1}} \dx + \co(\Delta_{m,n}).
\end{align*}

Recalling $\phi^{q}_{m-1,n}(x)-\phi(x)=\sum_{r=1}^{m-3}q_{r}(x)n^{-r/2}$, we get

\begin{align*}
\big(\phi^{q}_{m-1,n}(x)-\phi(x)\big)^k
=\sum_{l=1}^{k(m-3)} n^{-l/2} \sum q_{r_1}(x)\dots q_{r_k}(x)
\end{align*}

where the inner sum runs over all positive integers $r_1,\dots,r_k\le m-3$ with $r_1+\dots+r_k=l$. So

\begin{align*}
D(T_n)= \sum_{k=2}^{m-2} \frac{(-1)^k}{k(k-1)} \sum_{l=1}^{k(m-3)} n^{-l/2} \sum \int_{\R} \frac{q_{r_1}(x)\dots q_{r_k}(x)}{\phi(x)^{k-1}} \dx + \co(\Delta_{m,n}).
\end{align*}

Here, all summands with uneven $l=r_1+\dots+r_k$ vanish as $q_{r}=0$ for uneven $r$.
Additionally, all summands with $l>m-2$ will be absorbed by the remainder $\co(\Delta_{m,n})$ which yields

\begin{equation}\label{eq.D(Tn)-finish}
\begin{split}
&= \sum_{k=2}^{m-2} \frac{(-1)^k}{k(k-1)} \sum_{l=1,~l \text{ even}}^{m-2} n^{-l/2} \sum \int_{\R} \frac{q_{r_1}(x)\dots q_{r_k}(x)}{\phi(x)^{k-1}} \dx + \co(\Delta_{m,n})\\*
&= \sum_{k=2}^{m-2} \frac{(-1)^k}{k(k-1)} \sum_{l=1}^{\lfloor (m-2)/2 \rfloor} n^{-l} \sum \int_{\R} \frac{q_{r_1}(x)\dots q_{r_k}(x)}{\phi(x)^{k-1}} \dx + \co(\Delta_{m,n})
%	&= \sum_{l=1}^{\lfloor (m-2)/2 \rfloor} n^{-l} c_l + \co(\Delta_{m,n}),
\end{split}
\end{equation}

where the inner sum runs over all positive integers $r_1,\dots,r_k\le m-3$ such that $r_1+\dots+r_k=l$ in the first line and $r_1+\dots+r_k=2l$ in the second line. Define

\begin{align*}
&=\sum_{k=2}^{m-2} \frac{(-1)^k}{k(k-1)} \sum \int_{\R} \frac{q_{r_1}(x)\dots q_{r_k}(x)}{\phi(x)^{k-1}} \dx\\*
&=\sum_{k=2}^{2l} \frac{(-1)^k}{k(k-1)} \sum \int_{\R} \frac{q_{r_1}(x)\dots q_{r_k}(x)}{\phi(x)^{k-1}} \dx.
\end{align*}

Here, the inner sum in the first line runs over all positive integers $r_1,\dots,r_k\le m-3$ such that $r_1+\dots+r_k=2l$ and the the inner sum in the second line runs over all positive integers $r_1,\dots,r_k$ such that $r_1+\dots+r_k=2l$. The fact $2\le k\le 2l\le m-2$ implies the second identity.
Note that

\begin{equation*}
= \frac{(-1)^2}{2} \sum_{r_1+r_2=2} \int_{\R} \frac{q_{r_1}(x) q_{r_2}(x)}{\phi(x)} \dx
= \frac{1}{2} \int_{\R} \frac{q_1(x) q_1(x)}{\phi(x)} \dx
\end{equation*}

and thus, <ref> follows from eq.D(Tn)-finish.

§ RATE OF CONVERGENCE FOR EDGEWORTH EXPANSIONS IN THE CENTRAL LIMIT THEOREM IN TOTAL VARIATION DISTANCE FOR SELF-NORMALIZED SUMS

The main goal of this section is to prove <ref> by examining the total variation distance between $F_n$ and $\Phi^{Q}_{m,n}$. By eq.TV-L1, this is equivalent to bounding the $L^1$ norm of the difference of the corresponding densities.

Assume that $X_1$ is symmetric, the distribution of $X_1$ is non-singular, <ref> is satisfied and $\E|X_1|^{2m}<\infty$ for some $m\in\N$, $m\ge3$. Then there exists $N\in\N$ such that for all $n\ge N$, the statistics $T_{n}$ have densities $f_{n}$ that satisfy

\begin{equation}\label{eq.t-dichte-L1}
\|f_n-\phi^{q}_{m,n}\|_{L^1}= \co\big(n^{-(m-2)/2}\big).
\end{equation}

Together with <ref> and eq.TV-L1, <ref> implies <ref>. On our route to proving <ref>, we first show the following proposition. For any $h\in\R$, let $f(\lbullet+h)$ denote the function $x\mapsto f(x+h)$.

Assume that $X_1$ is symmetric, the distribution of $X_1$ is non-singular and $\E|X_1|^{s}<\infty$ for some ${s}\ge2$. Additionally, assume that <ref> is satisfied and there exists $p\in[1,\infty)$ and $\alpha\in\R$ with $0<\alpha\le c(s)$ such that $\|f_n\|_{L^p}\le \exp(n^\alpha)$ for all $n$ sufficiently large. Let $m=\lfloor {s} \rfloor$ and $a_n=\co(1)$ with polynomial order. Then

\begin{equation}\label{eq.p-dichte-Lp}
\|f_n-\phi^{q}_{m,n}\|_{L^p} \le \sup_{|h|\le a_n} \big\|f_n(\lbullet)-f_n(\lbullet+h)\big\|_{L^p} + \co\big( n^{-({s}-2)/2} \, (\log n)^{({s}+2m)/2} \big).
\end{equation}

Set $\beta_n= a_n^2\cdot n^{-\max\{2\alpha,(s-2)/2\}}$ and recall $f_n'$ from eq.f_n'-def. Now, <ref> yields

\begin{equation*}
\sup_{x\in\R} \, (1+|x|)^{m}|f_n'(x)-\phi^{q}_{m,n}(x)|= r_n
\end{equation*}

for $r_n=\co\big( n^{-({s}-2)/2} \, (\log n)^{({s}+2m)/2} \big)$.
By Minkowski's inequality and <ref>,

\begin{align*}
\|f_n-\phi^{q}_{m,n}\|_{L^p}
&\le \|f_n-f_n'\|_{L^p} + \|f_n'-\phi^{q}_{m,n}\|_{L^p}\\
&\le \|f_n-f_n'\|_{L^p} + \|(1+|\lbullet|)^{-m}\|_{L^p} \cdot r_n\\
%	&= \|f_n-f_n'\|_{L^p} + \big(\tfrac{2}{pm-1}\big)^{1/p} \cdot r_n\\
&= \|f_n-f_n'\|_{L^p} + \co\big( n^{-({s}-2)/2} \, (\log n)^{({s}+2m)/2} \big).
\end{align*}

Next, for $b_n:=\sqrt{\beta_n}\cdot n^\alpha \le a_n$ and by <cit.>,

\begin{align*}
\|f_n-f_n'\|_{L^p}
%&= \Big\|\int_{-\infty}^{\infty} \big(f_n(\lbullet)-f_n(\lbullet-y)\big) \phi_{\beta_n}(y)\dy\Big\|_{L^p}\\
&\le \Big\|\int_{-\infty}^{\infty} \big(f_n(\lbullet)-f_n(\lbullet-y)\big) \phi_{\beta_n}(y) \1_{\{|y|\le b_n\}} \dy\Big\|_{L^p}\\
&\quad+\Big\|\int_{-\infty}^{\infty} \big(f_n(\lbullet)-f_n(\lbullet-y)\big) \phi_{\beta_n}(y) \1_{\{|y|> b_n\}} \dy\Big\|_{L^p}\\
&\le \big\|\phi_{\beta_n}(\lbullet)\1_{\{|\lbullet|\le b_n\}}\big\|_{L^1} \cdot \sup_{|h|\le b_n} \big\|f_n(\lbullet)-f_n(\lbullet+h)\big\|_{L^p}\\
&\quad + \big\|\phi_{\beta_n}(\lbullet)\1_{\{|\lbullet|> b_n\}}\big\|_{L^1} \cdot \sup_{|h|> b_n} \big\|f_n(\lbullet)-f_n(\lbullet+h)\big\|_{L^p}.
%&= \Big( \int_{|y|\le b_n} \phi_{\beta_n}(y) \dy \Big) \cdot \sup_{|h|\le b_n} \big\|f_n(\lbullet)-f_n(\lbullet+h)\big\|_{L^p}\\
%&\quad + \Big( \int_{|y|>b_n} \phi_{\beta_n}(y) \dy \Big) \cdot \sup_{|h|> b_n} \big\|f_n(\lbullet)-f_n(\lbullet+h)\big\|_{L^p}.
\end{align*}

As $1-\Phi(x)\le \phi(x)/x$, for $n$ sufficiently large, the expression above is bounded by

\begin{align*}
\lefteqn{ \sup_{|h|\le b_n} \big\|f_n(\lbullet)-f_n(\lbullet+h)\big\|_{L^p}
+ 2\big(1-\Phi\big(b_n/\sqrt{\beta_n}\,\big)\big)\cdot 2 \|f_n\|_{L^p}}\quad\\
&= \sup_{|h|\le b_n} \big\|f_n(\lbullet)-f_n(\lbullet+h)\big\|_{L^p}
+ 2\big(1-\Phi(n^\alpha)\big)\cdot 2 \|f_n\|_{L^p}\\
&\le \sup_{|h|\le a_n} \big\|f_n(\lbullet)-f_n(\lbullet+h)\big\|_{L^p} + 4 \,n^{-\alpha} \exp(-n^{2\alpha}/2+n^\alpha) \\
&= \sup_{|h|\le a_n} \big\|f_n(\lbullet)-f_n(\lbullet+h)\big\|_{L^p} + \co\big( n^{-({s}-2)/2} \, (\log n)^{({s}+2m)/2} \big)
\end{align*}

in particular.

The term

\begin{align*}
\sup_{|h|\le a_n} \|f_n(\lbullet)-f_n(\lbullet+h)\|_{L^p}
\end{align*}

from eq.p-dichte-Lp is known as the integral (or $L^p$) modulus of continuity. There exists plentiful literature on bounds of this term if the Fourier transform of $f_n$ satisfies various kinds of tail bounds (see e.g. [21, 26, 44]).

Next, we derive a bound on the $L^1$ modulus of continuity of $f_n$.

Assume that $X_1$ is symmetric, the distribution of $X_1$ is non-singular, <ref> is satisfied and $\E|X_1|^{2m}<\infty$ for some $m\in\N$, $m\ge3$. Let $a_n=\co\big(n^{-(9m-17)/2}\big)$. Then there exists $N\in\N$ such that for all $n\ge N$, the statistics $T_{n}$ have densities $f_{n}$ that satisfy

\begin{align}\label{eq.p-L1-modulus}
\sup_{|h|\le a_n} \|f_n(\lbullet)-f_n(\lbullet+h)\|_{L^1} = \co\big(n^{-(m-2)/2}\big).
\end{align}

Analogously to the proof of <ref>, <ref> and $\E |X_1|^{2m}<\infty$ yield that there exists $N\in \N$ such that eq.fn-int-gn,eq.BR-19.2 hold for all $n\ge N$. Throughout the remainder of the proof assume $n\ge N$. Now eq.fn-int-gn implies that the self-normalized sums $T_{n}$ have densities $f_n$. We expand

\[
\le K_1+K_2+K_3
\]

as in eq.f_n-f_n-Z, where the terms $K_1$, $K_2$ and $K_3$ are introduced. This time, the $2m$-th moment is assumed to be finite. Therefore with $w=m$, eq.k1,eq.k2 yield

\[
K_1+K_2\le |h| c(m) n^{4m-8},
\]

and thus

\begin{align*}
&\le |h| c(m) n^{4m-8}
+ K_3\\
&= |h| c(m) n^{4m-8}\\
&\quad+ R_n' \int_0^\infty \big(1+\|\gamma_{x,n}(z)\|\big)^{-m} \big(n^{-1/2}z\big)^{1/2}\\
&\qquad\qquad\quad + \big(1+\|\gamma_{x-h,n}(z)\|\big)^{-m} \big(n^{-1/2}z\big)^{1/2}\dz
\end{align*}

for all $|x|<2\sqrt{n}$ and $|h|<1$. Here, $R_n'=\co\big(n^{-(m-2)/2}\big)$ and $\gamma_{x,n}(z)=\big(x(n^{-1/2}z)^{1/2}, \allowbreak z-n^{1/2}\big)$. By <ref>,

\begin{equation}\label{eq.L1-modulus-zerlegung}
\begin{split}
\lefteqn{\sup_{|h|\le a_n} \|f_n(\lbullet)-f_n(\lbullet+h)\|_{L^1}}\quad\\
&\le\sup_{|h|\le a_n} \int_{-2\sqrt{n}}^{2\sqrt{n}} |h| c(m) n^{4m-8}\dx\\
&\quad + \sup_{|h|\le a_n} \int_{-2\sqrt{n}}^{2\sqrt{n}} R_n' \int_0^\infty \big(1+\|\gamma_{x,n}(z)\|\big)^{-m} \big(n^{-1/2}z\big)^{1/2}\\
&\qquad\qquad\qquad\qquad\qquad+ \big(1+\|\gamma_{x-h,n}(z)\|\big)^{-m} \big(n^{-1/2}z\big)^{1/2}\dz\dx.
\end{split}
\end{equation}

Regarding the first summand,

\begin{equation}\label{eq.L1-K3}
\sup_{|h|\le a_n} \int_{-2\sqrt{n}}^{2\sqrt{n}} |h| c(m) n^{4m-8}\dx
\le c(m) a_n \, n^{(8m-15)/2}\\
= \co\big(n^{-(m-2)/2}\big).
\end{equation}

For the second summand, it is obviously impossible to attain a bound in $h$. Therefore, we have to find a bound in $n$.

For $m\ge3$, by Fubini's theorem

\begin{align}\label{eq.int-Z4}
%	\begin{split}
\lefteqn{\int_{-2\sqrt{n}}^{2\sqrt{n}} \int_0^\infty \big(1+\|\gamma_{x,n}(z)\|\big)^{-m} \big(n^{-1/2}z\big)^{1/2} \dz \dx}\quad\nonumber\\
&\le \int_{-2\sqrt{n}}^{2\sqrt{n}} \int_0^\infty \big(1+\|\gamma_{x,n}(z)\|^2\big)^{-3/2} \big(n^{-1/2}z\big)^{1/2} \dz \dx\\
%	&= \int_0^\infty \int_{-2\sqrt{n}}^{2\sqrt{n}} \big(1+n^{-1/2}zx^2+(z-n^{1/2})^2\big)^{-3/2} \big(n^{-1/2}z\big)^{1/2} \dx \dz\nonumber\\
&= \int_0^\infty \big(n^{-1/2}z\big)^{1/2} \int_{-2\sqrt{n}}^{2\sqrt{n}} \big(1+n^{-1/2}zx^2+(z-n^{1/2})^2\big)^{-3/2} \dx \dz.\nonumber
% Auch mit leichterer Abschätzung kommt am Ende nix besseres raus (siehe "Z_4 verbessern?")
%	\end{split}
\end{align}

Applying <ref> gives

\begin{equation}\label{eq.inner-int}
\begin{split}
\lefteqn{\int_{-2\sqrt{n}}^{2\sqrt{n}} \big(1+n^{-1/2}zx^2+(z-n^{1/2})^2\big)^{-3/2} \dx}\quad\\
&= 2\big(1+4n^{1/2}z+(z-n^{1/2})^2\big)^{-1/2} (1+(z-n^{1/2})^2)^{-1} 2 n^{1/2}
\end{split}
\end{equation}

such that we have to examine

\begin{align*}
%	\begin{split}
\int_0^\infty \big(n^{-1/2}z\big)^{1/2} \big(1+4n^{1/2}z+(z-n^{1/2})^2\big)^{-1/2} (1+(z-n^{1/2})^2)^{-1} n^{1/2} \dz.
%	\end{split}
\end{align*}

We split this integral into three regions that we treat separately. First,

\begin{align*}
%	\begin{split}
&\int_0^{\sqrt{n}/2} \big(n^{-1/2}z\big)^{1/2} \big(1+4n^{1/2}z+(z-n^{1/2})^2\big)^{-1/2} (1+(z-n^{1/2})^2)^{-1} n^{1/2} \dz\\
&\quad\le c(0) \int_0^{\sqrt{n}/2} 1 \cdot n^{-1/2} \, n^{-1} \, n^{1/2} \dz
%	&\quad= c(0) n^{-3/4} \int_0^{\sqrt{n}/2} z^{-1/2} \dz\\
\le c(0) n^{-1/2}.
%	\end{split}
\end{align*}

Second, by <ref>

\begin{align*}
%	\begin{split}
&\int_{\frac{1}{2}\sqrt{n}}^{2\sqrt{n}} \big(n^{-1/2}z\big)^{1/2} \big(1+4n^{1/2}z+(z-n^{1/2})^2\big)^{-1/2} (1+(z-n^{1/2})^2)^{-1} n^{1/2} \dz\\
&\quad\le c(0) \int_{\frac{1}{2}\sqrt{n}}^{2\sqrt{n}} \sqrt{2} \cdot n^{-1/2} \, (1+(z-n^{1/2})^2)^{-1} \, n^{1/2} \dz\\
&\quad= c(0) \int_{\frac{1}{2}\sqrt{n}}^{2\sqrt{n}} (1+(z-n^{1/2})^2)^{-1} \dz\le c(0).
%	\end{split}
\end{align*}

\begin{align*}
%	\begin{split}
&\int_{2\sqrt{n}}^\infty \big(n^{-1/2}z\big)^{1/2} \big(1+4n^{1/2}z+(z-n^{1/2})^2\big)^{-1/2} (1+(z-n^{1/2})^2)^{-1} n^{1/2} \dz\\
&\quad\le c(0) \int_{2\sqrt{n}}^\infty n^{-1/4} z^{1/2} \big(z-n^{1/2}\big)^{-1} \, \big(z-n^{1/2}\big)^{-2} \, n^{1/2} \dz\\
%&\quad= c(0) \, n^{1/4} \int_{2\sqrt{n}}^\infty z^{1/2} \big(z-n^{1/2}\big)^{-3} \dz\\
&\quad\le c(0) \, n^{1/4} \int_{2\sqrt{n}}^\infty z^{1/2} \big(z-z/2\big)^{-3} \dz\\
&\quad= c(0) \, n^{1/4} \int_{2\sqrt{n}}^\infty z^{-5/2} \dz= c(0) \, n^{-1/2}.
%	\end{split}
\end{align*}

Combining eq.int-Z4, eq.inner-int and the last three formulas yields

\begin{equation*}%\label{eq.int-Z4-erg}
%	\begin{split}
\int_{-2\sqrt{n}}^{2\sqrt{n}} \int_0^\infty \big(1+\|\gamma_{x,n}(z)\|\big)^{-m} \big(n^{-1/2}z\big)^{1/2} \dz \dx
\le c(0).
%	\end{split}
\end{equation*}

This remains valid if $x$ is replaced by $x-h$ for any $|h|\le a_n$ and thus

\begin{align*}
%	\begin{split}
& \sup_{|h|\le a_n} \int_{-2\sqrt{n}}^{2\sqrt{n}} R_n' \int_0^\infty \big(1+\|\gamma_{x,n}(z)\|\big)^{-m} \big(n^{-1/2}z\big)^{1/2}\\*
&\qquad\qquad\qquad\qquad+ \big(1+\|\gamma_{x-h,n}(z)\|\big)^{-m} \big(n^{-1/2}z\big)^{1/2}\dz \dx\\
&\quad\le R_n' \, c(0)
%	\end{split}
\end{align*}

which yields eq.p-L1-modulus when combined with eq.L1-modulus-zerlegung,eq.L1-K3.

<ref> implies <ref>.

Due to the form eq.phi^q-def and the factor $\phi$ in all expansion terms $q_r$, $\|\phi^{q}_{m,n}\|_{L^p}\le c(m)$ for all $p\in[1,\infty)$. Thus by eq.fn-phi-Lp, $\|f_n\|_{L^p}\le c(m)$ for all $p\in[1,\infty)$ and all $n$ sufficiently large.

Set ${s}=m+1/2$ and $a_n=\co\big(n^{-(9m-16)/2}\big)$ with polynomial order.

Now <ref> imply eq.t-dichte-L1.

The bounds in the $L^1$ limit theorem are by $\log n$ tighter than in the LLT. This is because the integral modulus of continuity used in the $L^1$ setting is easier to estimate than the pointwise Hölder-type bound which was the equivalent condition in the LLT setting. As this bound was used in the proof of the entropic CLT, the discrepancy of $\log n$ transfers there. The reason behind the factor $\log n$ is the appearance of different exponents in <ref>.

§ PROOFS, REMARKS AND AUXILIARY LEMMAS

§.§ Proof of Proposition <ref>

We calculate the expected values of the $\tlambda_{l,n}$ appearing in $\tP_{2,n}$ and $\tP_{4,n}$ (see eq.tP-2+4). For clear illustration, let all moments be finite in the following procedure. Additionally, we will explicitly write $\mu_2$ instead of its fixed value 1 as it helps to keep track of the order of moments. First, we need the expected values of the conditional cumulants divided by powers of $V_n$ for which we expand

\begin{equation}\label{eq.V_n^-2}
\begin{split}
%&=n^{-1}\Big(n^{-1} \sum X_j^2\Big)^{-1}\\
&=n^{-1}\Big(\mu_2+n^{-1} \Big(\sum (X_j^2-\mu_2)\Big)\Big)^{-1}\\
&= n^{-1} \mu_2^{-1}
- n^{-2} \mu_2^{-2} \Big(\sum (X_j^2-\mu_2)\Big) \\
&\quad+ n^{-3} \mu_2^{-3} \Big(\sum (X_j^2-\mu_2)\Big)^2
+ \cO_p\big(n^{-3}\big)
\end{split}
\end{equation}

\begin{align*}
%&= n^{-2} \mu_2^{-2}
%- n^{-3} 2 \mu_2^{-3} \Big(\sum (X_j^2-\mu_2)\Big)
%+ n^{-4} \bigg(2 \mu_2^{-4} \Big(\sum (X_j^2-\mu_2)\Big)^2 + \mu_2^{-4} \Big(\sum (X_j^2-\mu_2)\Big)^2 \bigg)
%+ \cO_p\big(n^{-4}\big)\\
&= n^{-2} \mu_2^{-2}
- n^{-3} 2 \mu_2^{-3} \Big(\sum (X_j^2-\mu_2)\Big)
+ n^{-4} 3 \mu_2^{-4} \Big(\sum (X_j^2-\mu_2)\Big)^2
+ \cO_p\big(n^{-4}\big),\\
%&= n^{-3} \mu_2^{-3}
%- n^{-4} 3 \mu_2^{-4} \Big(\sum (X_j^2-\mu_2)\Big)
%+ n^{-5} \bigg(3 \mu_2^{-5} \Big(\sum (X_j^2-\mu_2)\Big)^2 + 3 \mu_2^{-5} \Big(\sum (X_j^2-\mu_2)\Big)^2 \bigg)
%+ \cO_p\big(n^{-5}\big)\\
&= n^{-3} \mu_2^{-3}
- n^{-4} 3 \mu_2^{-4} \Big(\sum (X_j^2-\mu_2)\Big)
+ n^{-5} 6 \mu_2^{-5} \Big(\sum (X_j^2-\mu_2)\Big)^2
+ \cO_p\big(n^{-5}\big),\\
%&= n^{-4} \mu_2^{-4}
%- n^{-5} 4 \mu_2^{-5} \Big(\sum (X_j^2-\mu_2)\Big)
%+ n^{-6} \bigg(4 \mu_2^{-6} \Big(\sum (X_j^2-\mu_2)\Big)^2 + 6 \mu_2^{-6} \Big(\sum (X_j^2-\mu_2)\Big)^2 \bigg)
%+ \cO_p\big(n^{-6}\big)\\
&= n^{-4} \mu_2^{-4}
- n^{-5} 4 \mu_2^{-5} \Big(\sum (X_j^2-\mu_2)\Big)
+ n^{-6} 10 \mu_2^{-6} \Big(\sum (X_j^2-\mu_2)\Big)^2
+ \cO_p\big(n^{-6}\big).
\end{align*}

Due to $\E X_j^2=\mu_2$, the index of every factor $(X_j^2-\mu_2)$ within

\[
\Big(\sum (X_j^2-\mu_2)\Big)^{k}\sum|X_j|^{2 l}
=\sum_{j_1,\dots,j_{k+1}=1}^n \big(X_{j_1}^2-\mu_2\big) \cdots \big( X_{j_k}^2-\mu_2 \big) |X_{j_{k+1}}|^{2 l}
\]

has to be equal to the index of another factor or the summand vanishes within the expectation. Thus, for $k\in\N$, the expectation of summands with $\big(\sum (X_j^2-\mu_2)\big)^{2k-1}$ and ${\big(\sum (X_j^2-\mu_2)\big)^{2k}}$ produce the same order of $n$. This yields

\begin{align*}
\E\Big[V_n^{-4}\sum|X_j|^4\Big]
%n^{-2} \mu_2^{-2} \sum|X_j|^4
%- n^{-3} 2 \mu_2^{-3} \Big(\sum (X_j^2-\mu_2)\Big) \sum|X_j|^4
%+ n^{-4} 3 \mu_2^{-4} \Big(\sum (X_j^2-\mu_2)\Big)^2 \sum|X_j|^4
%+ \cO_p\big(n^{-3}\big)\Big]\\
%&=n^{-2} \mu_2^{-2} n \mu_4
%- n^{-3} 2 \mu_2^{-3} n (\mu_6-\mu_2\mu_4)
%+ n^{-4} 3 \mu_2^{-4} n^2 (\mu_4-\mu_2^2)\mu_4
%+ \co\big(n^{-2}\big)\\
&=n^{-1} \mu_2^{-2} \mu_4
- n^{-2} \mu_2^{-4} \big(2\mu_2\mu_6 +\mu_2^2\mu_4 - 3 \mu_4^2\big)
+ \cO\big(n^{-3}\big),\\
\E\Big[V_n^{-6}\sum|X_j|^6\Big]
%n^{-3} \mu_2^{-3} \sum|X_j|^6
%- n^{-4} 3 \mu_2^{-4} \Big(\sum (X_j^2-\mu_2)\Big) \sum|X_j|^6
%+ n^{-5} 6 \mu_2^{-5} \Big(\sum (X_j^2-\mu_2)\Big)^2 \sum|X_j|^6
%+ \cO_p\big(n^{-4}\big)\Big]\\
%&=n^{-3} \mu_2^{-3} n \mu_6
%- n^{-4} 3 \mu_2^{-4} n (\mu_8-\mu_6\mu_2)
%+ n^{-5} 6 \mu_2^{-5} n^2 (\mu_4-\mu_2^2)\mu_6
%+ \co\big(n^{-3}\big)\\
&=n^{-2} \mu_2^{-3} \mu_6
- n^{-3} \mu_2^{-5} (3\mu_8\mu_2+3\mu_6\mu_2^2-6\mu_4\mu_6)
+ \cO\big(n^{-4}\big),\\
\E\Big[V_n^{-8}\Big(\sum|X_j|^4\Big)^2\Big]
%n^{-4} \mu_2^{-4} \Big(\sum|X_j|^4\Big)^2
%- n^{-5} 4 \mu_2^{-5} \Big(\sum (X_j^2-\mu_2)\Big) \Big(\sum|X_j|^4\Big)^2
%+ \cO_p\big(n^{-4}\big)\Big]\\
%&=n^{-4} \mu_2^{-4} \Big(n^2\mu_4^2+n(\mu_8-\mu_4^2)\Big)
%- n^{-5} 4 \mu_2^{-5} 2n^2(\mu_6\mu_4-\mu_4^2\mu_2)
%+ \co\big(n^{-3}\big)\\
&=n^{-2} \mu_2^{-4} \mu_4^2
- n^{-3} \mu_2^{-5} (8\mu_6\mu_4-7\mu_4^2\mu_2- \mu_8\mu_2)
+ \cO\big(n^{-4}\big)
\end{align*}

such that by eq.tk-2r,eq.tlambda

\begin{align*}
\E\bigg[\frac{\tlambda_{4,n}}{4!}\bigg]
&= n^{-1} \big(- \tfrac{1}{12}\big) \mu_2^{-2} \mu_4
+ n^{-2} \tfrac{1}{12} \mu_2^{-4} \big(2\mu_2\mu_6 +\mu_2^2\mu_4 - 3 \mu_4^2\big)
+ \cO\big(n^{-3}\big),\\
\E\bigg[\frac{\tlambda_{6,n}}{6!}\bigg]
&= n^{-2} \tfrac{1}{45} \mu_2^{-3} \mu_6
+ n^{-3} \big(- \tfrac{1}{45}\big) \mu_2^{-5} (3\mu_8\mu_2+3\mu_6\mu_2^2-6\mu_4\mu_6)
+ \cO\big(n^{-4}\big),\\
\E\bigg[\frac{1}{2} \Big(\frac{\tlambda_{4,n}}{4!}\Big)^2\bigg]
&=n^{-2} \tfrac{1}{288} \mu_2^{-4} \mu_4^2
+ n^{-3} \big(- \tfrac{1}{288}\big) \mu_2^{-5} (8\mu_6\mu_4-7\mu_4^2\mu_2- \mu_8\mu_2)
+ \cO\big(n^{-4}\big).
\end{align*}

The connection $\E\big[\Phi^{\tP}_{m,n}\big]$ and $\Phi^Q_{m,n}$ (the procedure for $\E\big[\phi^{\tp}_{m,n}\big]$ and $\phi^q_{m,n}$ is the same) is given in eq.Phi^tP, eq.tP, eq.Q-def and eq.Phi^Q-def and the corresponding computations are conducted in <ref>. However, we need to address the issue of moments higher than ${s}$ (as they could be infinite) and the factor $\exp(x^2/4)$. For ${s}<4$, there are no expansion terms in eq.E-tP-Q,eq.E-tp-q and there is nothing to prove. So assume ${s}\ge4$.

As in <ref>, we first evaluate the expectation of $V_n^{-2\alpha}\sum|X_j|^{2\alpha}$ for $\alpha\in\N$, ${2\le \alpha\le m/2}$. Define ${h}=2\lceil \frac{m-2\alpha+1}{2}\rceil$. Similar to eq.V_n^-2, we perform a Taylor expansion of $V_n^{-2\alpha}$ up to order ${h}-1$ as follows

\begin{equation}\label{eq.Exp-tL}
\begin{split}
&=\sum_{r=0}^{{h}-1} (-1)^r n^{-(\alpha+r)} t_{r,2\alpha} \Big(\sum (X_j^2-1)\Big)^{r}\Big(\sum|X_j|^{2\alpha}\Big)\\
&\quad+(-1)^{{h}} n^{-(\alpha+{h})} t_{{h},2\alpha} \xi^{-(\alpha+{h})} \Big(\sum (X_j^2-1)\Big)^{{h}}\Big(\sum|X_j|^{2\alpha}\Big),
\end{split}
\end{equation}

with (random) intermediate value $\xi$ between $1$ and $n^{-1} V_n^2$. Here, $t_{r,2\alpha}$ are the constant factors which appear in this Taylor expansion and are bounded in absolute value by $c(m)$.

First, we will show that the sum of all terms where $X_\cdot$ appears with a power higher than ${s}$ is of order $\co \big(n^{-({s}-2)/2}\big)$. The same is shown for the remainder of the Taylor expansion including the $\xi$.

Note that by eq.Lle1,eq.Mn-delta,eq.Vn-bound-1/2, we can restrict the expectation of the terms in eq.Exp-tL to the event

\begin{equation}\label{eq.event}
A_{n,1}\cap A_{n,2} \quad \text{where} \quad A_{n,1}:=\{M_n< \delta_n \sqrt{n} \, \} \quad\text{and} \quad A_{n,2}:=\{V_n^2 > \tfrac{n}{2}\}
\end{equation}

with an error of order $\co \big(n^{-({s}-2)/2}\big)$. The sequence $(\delta_n)$ is introduced in <ref>.

\begin{equation}\label{eq.E-expanded}
\begin{split}
\lefteqn{\E\Big[ V_n^{-2\alpha}\sum|X_j|^{2\alpha} \Big]}\quad\\
&=\E\Big[ \1_{A_{n,1}\cap A_{n,2}} V_n^{-2\alpha}\Big(\sum|X_j|^{2\alpha}\Big) \Big] + \co \big(n^{-({s}-2)/2}\big)\\
&=\E\bigg[ \1_{A_{n,1}\cap A_{n,2}} \sum_{r=0}^{{h}-1} (-1)^r n^{-(\alpha+r)} t_{r,2\alpha} \Big(\sum (X_j^2-1)\Big)^{r}\Big(\sum|X_j|^{2\alpha}\Big)\bigg]\\
&\quad+\E\bigg[ \1_{A_{n,1}\cap A_{n,2}} (-1)^{{h}} n^{-(\alpha+{h})} t_{{h},2\alpha} \xi^{-(\alpha+{h})} \Big(\sum (X_j^2-1)\Big)^{{h}}\Big(\sum|X_j|^{2\alpha}\Big) \bigg] \\
&\quad+ \co \big(n^{-({s}-2)/2}\big).
\end{split}
\end{equation}

Note that by eq.event

\begin{align*}
%	&\bigg|\E\Big[\1_{A_{n,1}} \sum_{r=0}^{{h}-1} (-1)^r n^{-(\alpha+r)} t_{r,2\alpha} \Big(\sum (X_j^2-1)\Big)^{r}\Big(\sum|X_j|^{2\alpha}\Big)\Big]\\
%	&- \E\Big[\1_{A_{n,1}\cap A_{n,2}} \sum_{r=0}^{{h}-1} (-1)^r n^{-(\alpha+r)} t_{r,2\alpha} \Big(\sum (X_j^2-1)\Big)^{r}\Big(\sum|X_j|^{2\alpha}\Big)\Big]\bigg|\\
\lefteqn{\bigg|\E\bigg[\1_{A_{n,1}\cap A_{n,2}^c} \sum_{r=0}^{{h}-1} (-1)^r n^{-(\alpha+r)} t_{r,2\alpha} \Big(\sum (X_j^2-1)\Big)^{r}\Big(\sum|X_j|^{2\alpha}\Big)\bigg]\bigg|}\quad\\*
&\le c(h) \sum_{r=0}^{{h}-1} \E\Big[\1_{A_{n,1}\cap A_{n,2}^c} n^{-(\alpha+r)} \big(V_n^2+n\big)^{r} V_n^2 (\delta_n \sqrt{n} \, )^{2\alpha-2}\Big]\\
&\le c(h) \sum_{r=0}^{{h}-1} \E\Big[\1_{A_{n,2}^c} n^{-(r+1)} \big(\tfrac{n}{2}+n\big)^{r} \tfrac{n}{2}\Big] \\
&\le c(h) \P\big(A_{n,2}^c\big)\\
&= \co \big(n^{-({s}-2)/2}\big)
\end{align*}

and thus,

\begin{equation}\label{eq.A_{n,2}-raus}
\begin{split}
&\E\bigg[ \1_{A_{n,1}\cap A_{n,2}} \sum_{r=0}^{{h}-1} (-1)^r n^{-(\alpha+r)} t_{r,2\alpha} \Big(\sum (X_j^2-1)\Big)^{r}\Big(\sum|X_j|^{2\alpha}\Big)\bigg]\\
&= \sum_{r=0}^{{h}-1} (-1)^r n^{-(\alpha+r)} t_{r,2\alpha} \, \E\Big[ \1_{A_{n,1}} \Big(\sum (X_j^2-1)\Big)^{r}\Big(\sum|X_j|^{2\alpha}\Big)\Big]
+ \co \big(n^{-({s}-2)/2}\big).
\end{split}
\end{equation}

Initially, we examine

\[
n^{-(\alpha+r)} \, \E\Big[ \1_{A_{n,1}} \Big(\sum (X_j^2-1)\Big)^{r}\Big(\sum|X_j|^{2\alpha}\Big)\Big]
\]

for $r=0,\dots,m$.

We expand

\[
\Big(\sum (X_j^2-1)\Big)^{r}
=\sum_{i_1+\dots+i_{n}=r} \binom{r}{i_1,\dots,i_n} \prod (X_j^2-1)^{i_{j}}.
\]

Here the sum extends over all $n$-tuples $(i_1,...,i_n)$ of non-negative integers with $\sum_{j=1}^ni_j=r$ and

\[
{r \choose i_{1},\ldots ,i_{n}}={\frac {r!}{i_{1}!\,i_{2}!\cdots i_{n}!}}.
\]

Since $X_1,\dots,X_n$ are i.i.d.,

\begin{align*}
\E\bigg[ {r \choose i_{\pi(1)},\ldots ,i_{\pi(n)}}\prod (X_j^2-1)^{i_{\pi(j)}} \bigg]
=	\E\bigg[ {r \choose i_{1},\ldots ,i_{n}}\prod (X_j^2-1)^{i_{j}}	\bigg]
\end{align*}

for all (non-random) permutations $\pi$ from $\{1,\dots,n\}$ to itself.

For $l=1,\dots,r$ let $k_l$ be the number of appearances of $l$ in the tuple $(i_1,...,i_n)$ (so $\sum_{l=1}^{r}lk_l =\sum_{j=1}^ni_j=r$). Then for each tuple $(k_1,\dots,k_r)$, there exist ${n \choose k_{1},\ldots ,k_{r},(n-\sum_{l=1}^{r} k_l)}$ tuples $(i_1,...,i_n)$ such that $k_l$ is the number of appearances of $l$ in the tuple $(i_1,...,i_n)$. So instead of summing over all $n$-tuples $(i_1,...,i_n)$ with $\sum_{j=1}^ni_j=r$, we can sum over all $r$-tuples $(k_1,...,k_r)$ of non-negative integers with $\sum_{l=1}^{r}lk_l =r$ and multiply each summand by ${n \choose k_{1},\ldots ,k_{r},(n-\sum_{l=1}^{r} k_l)}$. We perform the just described summation by using the sum $\sum_{*(k_\cdot,r,\cdot)}$ and $u(k_\cdot)$ which are explained in <ref>. Additionally, we convert

\[
{r \choose i_{1},\ldots ,i_{n}}={\frac {r!}{i_{1}!\,i_{2}!\cdots i_{n}!}}=\frac{r!}{1!^{k_1}\cdots r!^{k_r}}.
\]

Now for any $(k_1,\dots,k_r)$, we identify that element of the corresponding set

\[
\{(i_{\pi(1)},...,i_{\pi(n)}): \pi \text{ permutations from }\{1,\dots,n\}\text{ to itself} \}
\]

for which

\begin{align*}
\end{align*}

When summing over $(k_1,\dots,k_r)$, this will be the representative, evaluated within the expectation. This reordering yields

\begin{align*}
&\E\left[\Big(\sum (X_j^2-1)\Big)^{r} \Big(\sum|X_j|^{2\alpha}\Big) \right]\\
&\quad=\E\left[\sum_{i_1+\dots+i_{n}=r} \binom{r}{i_1,\dots,i_n} \Big(\prod (X_j^2-1)^{i_{j}}\Big) \Big(\sum|X_j|^{2\alpha}\Big) \right]\\
&\quad=\E\Bigg[\sum_{*(k_\cdot,r,\cdot)} {n \choose k_{1},\ldots ,k_{r},(n-u(k_\cdot))} \frac{r!}{1!^{k_1}\cdots r!^{k_r}}\\
&\quad\qquad\cdot\bigg( \prod_{l=1}^{r} \prod_{a=1}^{k_l}\Big(X_{\sum_{b=1}^{l-1}k_b+a}^2-1 \Big)^{l} \bigg) \Big(\sum|X_j|^{2\alpha}\Big) \Bigg].
\end{align*}

As $M_n$ is invariant under permutation, the last set of formulas remains valid if $\1_{A_{n,1}}$ is present in the expectation. So

\begin{equation}\label{eq.expanded}
\begin{split}
&n^{-(\alpha+r)} \, \E\Big[ \1_{A_{n,1}} \Big(\sum (X_j^2-1)\Big)^{r}\Big(\sum|X_j|^{2\alpha}\Big)\Big]\\
&\quad= n^{-(\alpha+r)} \sum_{*(k_\cdot,r,\cdot)} {n \choose k_{1},\ldots ,k_{r},(n-u(k_\cdot))} \frac{r!}{1!^{k_1}\cdots r!^{k_r}}\\
&\quad\qquad\cdot \sum \E\bigg[ \1_{A_{n,1}} \bigg( \prod_{l=1}^{r} \prod_{a=1}^{k_l}\Big(X_{\sum_{b=1}^{l-1}k_b+a}^2-1 \Big)^{l} \bigg) |X_j|^{2\alpha} \bigg]
\end{split}
\end{equation}

for all $r=0,\dots,m$.

Below, we distinguish between the values of $j$, which only coincide with the index of another factor for $j\le u(k_\cdot)$, and get

\begin{equation}\label{eq.j-split}
\begin{split}
&\sum \E\bigg[ \1_{A_{n,1}} \bigg( \prod_{l=1}^{r} \prod_{a=1}^{k_l}\Big(X_{\sum_{b=1}^{l-1}k_b+a}^2-1 \Big)^{l} \bigg) |X_j|^{2\alpha} \bigg]\\
&\quad= \sum_{j=u(k_\cdot)+1}^{n} \E\bigg[ \1_{A_{n,1}} \bigg( \prod_{l=1}^{r} \prod_{a=1}^{k_l}\Big(X_{\sum_{b=1}^{l-1}k_b+a}^2-1 \Big)^{l} \bigg) |X_j|^{2\alpha} \bigg]\\
&\quad\quad+ \sum_{j=1}^{u(k_\cdot)} \E\bigg[ \1_{A_{n,1}} \bigg( \prod_{l=1}^{r} \prod_{a=1}^{k_l}\Big(X_{\sum_{b=1}^{l-1}k_b+a}^2-1 \Big)^{l} \bigg) |X_j|^{2\alpha} \bigg].
\end{split}
\end{equation}

Next, we use that $X_1,\dots,X_n$ are i.i.d. and

\begin{equation}\label{eq.A_{n,1}-exp}
A_{n,1}=\{|X_1|<\delta_n \sqrt{n} \, \} \cap \dots \cap \{|X_n|<\delta_n \sqrt{n} \, \}.
\end{equation}

Additionally, we introduce the notation

\[
\E_{\delta_n,n}[\,\cdot\,]:=\E[\1_{\{|X_1|<\delta_n \sqrt{n} \, \}}\,\cdot\,].
\]

For the upper part of the sum this yields

\begin{align*}
&\sum_{j=u(k_\cdot)+1}^{n} \E\bigg[ \1_{A_{n,1}} \bigg( \prod_{l=1}^{r} \prod_{a=1}^{k_l}\Big(X_{\sum_{b=1}^{l-1}k_b+a}^2-1 \Big)^{l} \bigg) |X_j|^{2\alpha} \bigg]\\*
&= \big(n-u(k_\cdot)\big) \bigg( \prod_{l=1}^{r}\Big(\E_{\delta_n,n}\Big[ \big(X_1^2-1 \big)^{l} \Big]\Big)^{k_l} \bigg) \E_{\delta_n,n}\Big[ |X_1|^{2\alpha} \Big] \P\big(\{|X_1|<\delta_n \sqrt{n} \, \}\big)^{n-u(k_\cdot)-1}.
\end{align*}

Assume that $k_l>0$ for at least one $l \ge \lfloor s/2\rfloor+1$ (i.e. $2l>{s}$). Furthermore, in the following formula we use the fact $2\le\alpha\le m/2$. As

\[
{n \choose k_{1},\ldots ,k_{r},(n-u(k_\cdot))}\le c(r) n^{u(k_\cdot)}
\qquad
\text{and}
\qquad
\big(n-u(k_\cdot)\big)\le n,
\]

these summands of $\sum_{*(k_\cdot,r,\cdot)}$ in eq.expanded can be bounded by

\begin{align*}
&c(r) n^{-\alpha-r+u(k_\cdot)+1}
\bigg( \prod_{l=1}^{r}\Big(\E_{\delta_n,n}\Big[ \big(X_1^2-1 \big)^{l} \Big]\Big)^{k_l} \bigg)\\
& \qquad \cdot \E_{\delta_n,n}\Big[ |X_1|^{2\alpha} \Big] \P\big(\{|X_1|<\delta_n \sqrt{n} \, \}\big)^{n-u(k_\cdot)-1} \\
&=c(r) n^{1-\alpha}
\bigg( \prod_{l=1}^{r}\Big(n^{1-l}\E_{\delta_n,n}\Big[ \big(X_1^2-1 \big)^{l} \Big]\Big)^{k_l} \bigg)\\
& \qquad \cdot \E_{\delta_n,n}\Big[ |X_1|^{2\alpha} \Big] \P\big(\{|X_1|<\delta_n \sqrt{n} \, \}\big)^{n-u(k_\cdot)-1} \\
&=c(r) n^{1-\alpha}
\bigg( \prod_{l=1}^{\lfloor s/2\rfloor}\Big(n^{1-l}\E_{\delta_n,n}\Big[ \big(X_1^2-1 \big)^{l} \Big]\Big)^{k_l} \bigg) \bigg( \prod_{l=\lfloor s/2\rfloor+1}^{r}\Big(n^{1-l}\E_{\delta_n,n}\Big[ \big(X_1^2-1 \big)^{l} \Big]\Big)^{k_l} \bigg) \\
& \qquad \cdot \E_{\delta_n,n}\Big[ |X_1|^{2\alpha} \Big] \P\big(\{|X_1|<\delta_n \sqrt{n} \, \}\big)^{n-u(k_\cdot)-1} \\
&\le c(r) n^{1-\alpha}
\bigg( \prod_{l=1}^{\lfloor s/2\rfloor}\Big(n^{1-l}\E_{\delta_n,n}\Big[ \big(X_1^2-1 \big)^{l} \Big]\Big)^{k_l} \bigg) \\
& \qquad \cdot \bigg( \prod_{l=\lfloor s/2\rfloor+1}^{r}\Big(n^{1-\lfloor s/2\rfloor}\delta_n^{2(l-\lfloor s/2\rfloor)} \E_{\delta_n,n}\Big[ \big(X_1^2-1 \big)^{\lfloor s/2\rfloor} \Big]\Big)^{k_l} \bigg)\\
& \qquad \cdot \E_{\delta_n,n}\Big[ |X_1|^{2\alpha} \Big] \P\big(\{|X_1|<\delta_n \sqrt{n} \, \}\big)^{n-u(k_\cdot)-1} \\
&\le c(r) n^{1-\alpha+1-\lfloor s/2\rfloor} \delta_n^2 \\
&= \co \big(n^{-({s}-2)/2}\big).
\end{align*}

Thus, the upper part of the sum reduces to

\begin{equation}\label{eq.j-ge}
\begin{split}
%	& n^{-(\alpha+r)} \sum_{*(k_\cdot,r,\cdot)} {n \choose k_{1},\ldots ,k_{r},(n-u(k_\cdot))} \frac{r!}{1!^{k_1}\cdots r!^{k_r}}\\
%	&\qquad \sum_{j=u(k_\cdot)+1}^{n} \E\bigg[ \1_{A_{n,1}} \bigg( \prod_{l=1}^{r} \prod_{a=1}^{k_l}\Big(X_{\sum_{b=1}^{l-1}k_b+a}^2-1 \Big)^{l} \bigg) |X_j|^{2\alpha} \bigg]\\
& n^{-(\alpha+r)} \sum_{*(k_\cdot,r,\cdot)} {n \choose k_{1},\ldots ,k_{r},(n-u(k_\cdot))} \frac{r!}{1!^{k_1}\cdots r!^{k_r}} \big(n-u(k_\cdot)\big) \1_{\{k_{\lfloor s/2\rfloor+1}=\dots=k_r=0\}}\\
&\qquad \cdot \bigg( \prod_{l=1}^{r}\Big(\E_{\delta_n,n}\Big[ \big(X_1^2-1 \big)^{l} \Big]\Big)^{k_l} \bigg) \E_{\delta_n,n}\Big[ |X_1|^{2\alpha} \Big] \P\big(\{|X_1|<\delta_n \sqrt{n} \, \}\big)^{n-u(k_\cdot)-1}\\
&\quad+\co \big(n^{-({s}-2)/2}\big).
\end{split}
\end{equation}

Now to the lower part of the sum in eq.j-split. We treat the summands separately. For any $j$, define $l^*=l^*(j)\in\{1,\dots,r\}$ as the power of the term $(X_{j}^2-1 )$ within the products in eq.j-split. in turn, for any $l^*$, let $j^*$ be the smallest index with the power $l^*$, i.e. $j^*=j^*(l^*):=\argmin\{j:l^*(j)=l^*\}$. With this assignment $j^*=\sum_{b=1}^{{l^*}-1}k_b+1$.

We use that $X_1,\dots,X_n$ are i.i.d. and eq.A_n,1-exp, which yields

\begin{align*}
\lefteqn{\E\bigg[ \1_{A_{n,1}} \bigg( \prod_{l=1}^{r} \prod_{a=1}^{k_l}\Big(X_{\sum_{b=1}^{l-1}k_b+a}^2-1 \Big)^{l} \bigg) |X_j|^{2\alpha} \bigg]}\quad\\*
&=\E\bigg[ \1_{A_{n,1}} \bigg( \prod_{l=1}^{r} \prod_{a=1}^{k_l}\Big(X_{\sum_{b=1}^{l-1}k_b+a}^2-1 \Big)^{l} \bigg) |X_{\sum_{b=1}^{{l^*}-1}k_b+1}|^{2\alpha} \bigg]\\
&= \bigg( \prod_{l=1 , ~ l \ne l^*}^{r}\Big(\E_{\delta_n,n}\Big[ \big(X_1^2-1 \big)^{l} \Big]\Big)^{k_l} \bigg) \Big(\E_{\delta_n,n}\Big[ \big(X_1^2-1 \big)^{l^*} \Big]\Big)^{k_{l^*}-1}\\
& \qquad \cdot \E_{\delta_n,n}\Big[ \big(X_1^2-1 \big)^{l^*}|X_1|^{2\alpha} \Big] \P\big(\{|X_1|<\delta_n \sqrt{n} \, \}\big)^{n-u(k_\cdot)}.
\end{align*}

Let $2\alpha+2l^*>{s}$. These summands of $\sum_{*(k_\cdot,r,\cdot)}$ in eq.expanded can be bounded by

\begin{align*}
&c(r) n^{-\alpha-r+u(k_\cdot)}
\bigg( \prod_{l=1 , ~ l \ne l^*}^{r}\Big(\E_{\delta_n,n}\Big[ \big(X_1^2-1 \big)^{l} \Big]\Big)^{k_l} \bigg) \Big(\E_{\delta_n,n}\Big[ \big(X_1^2-1 \big)^{l^*} \Big]\Big)^{k_{l^*}-1}\\*
& \qquad \cdot \E_{\delta_n,n}\Big[ \big(X_1^2-1 \big)^{l^*}|X_1|^{2\alpha} \Big] \P\big(\{|X_1|<\delta_n \sqrt{n} \, \}\big)^{n-u(k_\cdot)} \\
&=c(r) n^{-\alpha}
\bigg( \prod_{l=1 , ~ l \ne l^*}^{r}\Big(n^{1-l}\E_{\delta_n,n}\Big[ \big(X_1^2-1 \big)^{l} \Big]\Big)^{k_l} \bigg) \Big(n^{1-l^*}\E_{\delta_n,n}\Big[ \big(X_1^2-1 \big)^{l^*} \Big]\Big)^{k_{l^*}-1}\\
& \qquad \cdot n^{1-l^*} \E_{\delta_n,n}\Big[ \big(X_1^2-1 \big)^{l^*}|X_1|^{2\alpha} \Big] \P\big(\{|X_1|<\delta_n \sqrt{n} \, \}\big)^{n-u(k_\cdot)} \\
&\le c(r) n^{-\alpha}
\bigg( \prod_{l=1 , ~ l \ne l^*}^{r}\Big(\delta_n^{2(l-1)}\E_{\delta_n,n}\Big[ \big(X_1^2-1 \big) \Big]\Big)^{k_l} \bigg) \Big(\delta_n^{2(l^*-1)}\E_{\delta_n,n}\Big[ \big(X_1^2-1 \big) \Big]\Big)^{k_{l^*}-1}\\
& \qquad \cdot n^{1-l^*} \E_{\delta_n,n}\Big[ \big(|X_1|^{2l^*+2\alpha}+|X_1|^{2\alpha} \big) \Big] \P\big(\{|X_1|<\delta_n \sqrt{n} \, \}\big)^{n-u(k_\cdot)} \\
&\le c(r) n^{-\alpha}
\bigg( \prod_{l=1 , ~ l \ne l^*}^{r}\Big(\delta_n^{2(l-1)}\E_{\delta_n,n}\Big[ \big(X_1^2-1 \big) \Big]\Big)^{k_l} \bigg) \Big(\delta_n^{2(l^*-1)}\E_{\delta_n,n}\Big[ \big(X_1^2-1 \big) \Big]\Big)^{k_{l^*}-1}\\
& \qquad \cdot n^{1-l^*} (\delta_n \sqrt{n} \, )^{2l^*+2\alpha-{s}} \E_{\delta_n,n}\Big[ \big(|X_1|^{s}+|X_1|^{{s}-2l^*} \big) \Big] \P\big(\{|X_1|<\delta_n \sqrt{n} \, \}\big)^{n-u(k_\cdot)} \\
&\le c(r) n^{-\alpha+1-l^*+l^*+\alpha-{s}/2} \delta_n^{2l^*+2\alpha-{s}} \\
&= \co \big(n^{-({s}-2)/2} \big)
\end{align*}

since $\delta_n\to0$.

If $2\alpha+2l^*\le{s}$, a similar procedure and bound as for the upper part of the sum in eq.j-split apply. To this end, assume that $k_l>0$ for at least one $l \ge \lfloor s/2\rfloor+1$. Furthermore, in the following formula we use the fact $2\le\alpha\le m/2$. With $l^*$ as above, these summands of eq.expanded can be bounded by

\begin{align*}
&c(r) n^{-\alpha-r+u(k_\cdot)}
\bigg( \prod_{l=1 , ~ l \ne l^*}^{r}\Big(\E_{\delta_n,n}\Big[ \big(X_1^2-1 \big)^{l} \Big]\Big)^{k_l} \bigg) \Big(\E_{\delta_n,n}\Big[ \big(X_1^2-1 \big)^{l^*} \Big]\Big)^{k_{l^*}-1}\\*
& \quad \cdot \E_{\delta_n,n}\Big[ \big(X_1^2-1 \big)^{l^*}|X_1|^{2\alpha} \Big] \P\big(\{|X_1|<\delta_n \sqrt{n} \, \}\big)^{n-u(k_\cdot)} \\*
&=c(r) n^{-\alpha}
\bigg( \prod_{l=1 , ~ l \ne l^*}^{r}\Big(n^{1-l}\E_{\delta_n,n}\Big[ \big(X_1^2-1 \big)^{l} \Big]\Big)^{k_l} \bigg) \Big(n^{1-l^*}\E_{\delta_n,n}\Big[ \big(X_1^2-1 \big)^{l^*} \Big]\Big)^{k_{l^*}-1}\\
& \quad \cdot n^{1-l^*} \E_{\delta_n,n}\Big[ \big(X_1^2-1 \big)^{l^*}|X_1|^{2\alpha} \Big] \P\big(\{|X_1|<\delta_n \sqrt{n} \, \}\big)^{n-u(k_\cdot)} \\
&=c(r) n^{-\alpha}
\bigg( \prod_{l=1 , ~ l \ne l^*}^{\lfloor s/2\rfloor}\Big(n^{1-l}\E_{\delta_n,n}\Big[ \big(X_1^2-1 \big)^{l} \Big]\Big)^{k_l} \bigg) \Big(n^{1-l^*}\E_{\delta_n,n}\Big[ \big(X_1^2-1 \big)^{l^*} \Big]\Big)^{k_{l^*}-1}\\
& \quad \cdot n^{1-l^*} \E_{\delta_n,n}\Big[ \big(X_1^2-1 \big)^{l^*}|X_1|^{2\alpha} \Big] \bigg( \prod_{l=\lfloor s/2\rfloor+1}^{r}\Big(n^{1-l}\E_{\delta_n,n}\Big[ \big(X_1^2-1 \big)^{l} \Big]\Big)^{k_l} \bigg)\\
& \quad \cdot \P\big(\{|X_1|<\delta_n \sqrt{n} \, \}\big)^{n-u(k_\cdot)} \\
&\le c(r) n^{-\alpha}
\bigg( \prod_{l=1 , ~ l \ne l^*}^{\lfloor s/2\rfloor}\Big(n^{1-l}\E_{\delta_n,n}\Big[ \big(X_1^2-1 \big)^{l} \Big]\Big)^{k_l} \bigg) \Big(n^{1-l^*}\E_{\delta_n,n}\Big[ \big(X_1^2-1 \big)^{l^*} \Big]\Big)^{k_{l^*}-1}\\
& \quad \cdot n^{1-l^*} \E_{\delta_n,n}\Big[ \big(X_1^2-1 \big)^{l^*}|X_1|^{2\alpha} \Big] \\
&\quad \cdot \bigg( \prod_{l=\lfloor s/2\rfloor+1}^{r}\Big(n^{1-\lfloor s/2\rfloor}\delta_n^{2(l-\lfloor s/2\rfloor)} \E_{\delta_n,n}\Big[ \big(X_1^2-1 \big)^{\lfloor s/2\rfloor} \Big]\Big)^{k_l} \bigg) \P\big(\{|X_1|<\delta_n \sqrt{n} \, \}\big)^{n-u(k_\cdot)} \\
&\le c(r) n^{-\alpha+1-l^*+1-\lfloor s/2\rfloor} \delta_n^2 \\
&= \co \big(n^{-({s}-2)/2}\big).
\end{align*}

Summarizing, for the summands of $\sum_{*(k_\cdot,r,\cdot)}$ in eq.expanded with $j\le u(k_\cdot)$ we get

\begin{align}\label{eq.j-le}
%	\begin{split}
& n^{-(\alpha+r)} {n \choose k_{1},\ldots ,k_{r},(n-u(k_\cdot))} \frac{r!}{1!^{k_1}\cdots r!^{k_r}}\nonumber\\
&\quad \qquad \cdot \E\bigg[ \1_{A_{n,1}} \bigg( \prod_{l=1}^{r} \prod_{a=1}^{k_l}\Big(X_{\sum_{b=1}^{l-1}k_b+a}^2-1 \Big)^{l} \bigg) |X_j|^{2\alpha} \bigg]\nonumber\\
&\quad = n^{-(\alpha+r)} {n \choose k_{1},\ldots ,k_{r},(n-u(k_\cdot))} \frac{r!}{1!^{k_1}\cdots r!^{k_r}} \1_{\{k_{\lfloor s/2\rfloor+1}=\dots=k_r=0,\, 2\alpha+2l^*\le{s}\}}\\
&\quad \qquad \cdot \bigg( \prod_{l=1 , ~ l \ne l^*}^{r}\Big(\E_{\delta_n,n}\Big[ \big(X_1^2-1 \big)^{l} \Big]\Big)^{k_l} \bigg) \Big(\E_{\delta_n,n}\Big[ \big(X_1^2-1 \big)^{l^*} \Big]\Big)^{k_{l^*}-1}\nonumber\\
&\quad \qquad \cdot \E_{\delta_n,n}\Big[ \big(X_1^2-1 \big)^{l^*}|X_1|^{2\alpha} \Big] \P\big(\{|X_1|<\delta_n \sqrt{n} \, \}\big)^{n-u(k_\cdot)}\nonumber\\
&\quad\quad+\co \big(n^{-({s}-2)/2}\big).\nonumber
% + \1_{\{k_{\lfloor s/2\rfloor+1}=\dots=k_r=0,\, 2\alpha+2l^*>{s}\}} \cO \big(n^{-({s}-2)/2} \,\delta_n^{2l^*+2\alpha-{s}}\big).
%	\end{split}
\end{align}

Putting together eq.j-ge,eq.j-le,eq.j-split,eq.expanded, we get

\begin{align}\label{eq.together-indicator}
%	\begin{split}
&n^{-(\alpha+r)} \, \E\Big[ \1_{A_{n,1}} \Big(\sum (X_j^2-1)\Big)^{r}\Big(\sum|X_j|^{2\alpha}\Big)\Big]\\
&=n^{-(\alpha+r)} \sum_{*(k_\cdot,r,\cdot)} {n \choose k_{1},\ldots ,k_{r},(n-u(k_\cdot))} \frac{r!}{1!^{k_1}\cdots r!^{k_r}} \big(n-u(k_\cdot)\big) \1_{\{k_{\lfloor s/2\rfloor+1}=\dots=k_r=0\}}\nonumber\\*
&\qquad \cdot \bigg( \prod_{l=1}^{r}\Big(\E_{\delta_n,n}\Big[ \big(X_1^2-1 \big)^{l} \Big]\Big)^{k_l} \bigg) \E_{\delta_n,n}\Big[ |X_1|^{2\alpha} \Big] \P\big(\{|X_1|<\delta_n \sqrt{n} \, \}\big)^{n-u(k_\cdot)-1}\nonumber\\
& \quad + n^{-(\alpha+r)} \sum_{*(k_\cdot,r,\cdot)} {n \choose k_{1},\ldots ,k_{r},(n-u(k_\cdot))} \frac{r!}{1!^{k_1}\cdots r!^{k_r}} \nonumber\\
& \qquad \cdot \Bigg( \sum_{j=1}^{u(k_\cdot)} \1_{\{k_{\lfloor s/2\rfloor+1}=\dots=k_r=0,\, 2\alpha+2l^*\le{s}\}} \bigg( \prod_{l=1 , ~ l \ne l^*}^{r}\Big(\E_{\delta_n,n}\Big[ \big(X_1^2-1 \big)^{l} \Big]\Big)^{k_l} \bigg) \nonumber\\
& \qquad \cdot \Big(\E_{\delta_n,n}\Big[ \big(X_1^2-1 \big)^{l^*} \Big]\Big)^{k_{l^*}-1} \E_{\delta_n,n}\Big[ \big(X_1^2-1 \big)^{l^*}|X_1|^{2\alpha} \Big] \P\big(\{|X_1|<\delta_n \sqrt{n} \, \}\big)^{n-u(k_\cdot)} \Bigg)\nonumber\\*
& \quad +\co \big(n^{-({s}-2)/2}\big).\nonumber
%	\end{split}
\end{align}

Note that for $a=0,\dots, \lfloor s/2\rfloor$ and $b=0,\dots, \lfloor s/2\rfloor$ with $2a+2b\le{s}$, the expectations including the complementary indicator are bounded by

\begin{align}\label{eq.X_1^a*X_1^b-ge}
%	\begin{split}
\lefteqn{\bigg|\E\Big[ \1_{\{|X_1|\ge\delta_n \sqrt{n} \, \}}\big(X_1^2-1 \big)^{a} |X_1|^{2b} \Big] \bigg|}\quad\nonumber\\*
&\le \E\Big[ \1_{\{|X_1|\ge\delta_n \sqrt{n} \, \}}\big(|X_1|^{2a}+1 \big) |X_1|^{2b} \Big] \nonumber\\
&\le \E\Big[ \1_{\{|X_1|\ge\delta_n \sqrt{n} \, \}} (\delta_n \sqrt{n} \, )^{-(s-2a-2b)} \big(|X_1|^s+|X_1|^{s-2a} \big) \Big] \\
&\le c(0) n^{-(s/2-a-b)} \, \E\Big[ \1_{\{|X_1|\ge\delta_n \sqrt{n} \, \}} \delta_n ^{-s} |X_1|^s \Big] \nonumber\\
%	\end{split}%\co wegen EX^s<\infty und indikator wird kleiner
\end{align}

by eq.delta_n-def. In particular for $a=b=0$, $\P\big(\{|X_1|\ge\delta_n \sqrt{n} \, \}\big)=\co\big(n^{-s/2}\big)$.

For any $l_0=0,\dots,\lfloor s/2\rfloor$, the presence of a factor $\E_{\delta_n,n}\big[ \big(X_1^2-1 \big)^{l_0} \big]$ in a summand of eq.together-indicator implies $k_{l_0}\ge1$. Transitioning from $\E_{\delta_n,n}\big[ \big(X_1^2-1 \big)^{l_0} \big]$ to $\E\big[\big(X_1^2-1 \big)^{l_0} \big]$ in one of its appearances leads us to study the additional summand in eq.together-indicator

\begin{align*}
& n^{-(\alpha+r)} {n \choose k_{1},\ldots ,k_{r},(n-u(k_\cdot))} \frac{r!}{1!^{k_1}\cdots r!^{k_r}} \big(n-u(k_\cdot)\big) \1_{\{k_{\lfloor s/2\rfloor+1}=\dots=k_r=0\}}\\
&\qquad \cdot \bigg( \prod_{l=1,~l\ne l_0}^{r}\Big(\E_{\delta_n,n}\Big[ \big(X_1^2-1 \big)^{l} \Big]\Big)^{k_l} \bigg) \E_{\delta_n,n}\Big[ |X_1|^{2\alpha} \Big] \P\big(\{|X_1|<\delta_n \sqrt{n} \, \}\big)^{n-u(k_\cdot)-1}\\
&\qquad \cdot \Big(\E_{\delta_n,n}\Big[ \big(X_1^2-1 \big)^{l_0} \Big]\Big)^{k_{l_0}-1}
\bigg(- \E\Big[ \1_{\{|X_1|\ge\delta_n \sqrt{n} \, \}} \big(X_1^2-1 \big)^{l_0} \Big]\bigg)
\end{align*}

in the upper part and in the lower part correspondingly.

Using eq.X_1^a*X_1^b-ge, this summand is bounded in absolute value by

\begin{align*}
&c(m) n^{-(\alpha+r)} n^{u(k_\cdot)} n
\E\Big[ \1_{\{|X_1|\ge\delta_n \sqrt{n} \, \}} \big(X_1^2-1 \big)^{l_0} \Big]\\
&\quad\le c(m) n^{-\alpha-r+u(k_\cdot)+1}
\co\big(n^{-(s/2-l_0)}\big)\\
&\quad= \co\big(n^{-\alpha-r+u(k_\cdot)+1-s/2+l_0}\big).
\end{align*}

Using the facts $\alpha\ge 2$ and $k_{l_0}\ge1$, we examine the power of $n$

\begin{align*}
&\le -2-r+u(k_\cdot)+1-s/2+(l_0-1)k_{l_0}+1\\
&= -s/2-r+(k_1+\dots+k_{l_0}+\dots+k_r)+(l_0-1)k_{l_0}\\
&= -s/2-r+(k_1+\dots+l_0k_{l_0}+\dots+k_r)\\
&\le -s/2.
\end{align*}

Thus, leaving out the indicator in the factor $\E_{\delta_n,n}\big[ \big(X_1^2-1 \big)^{l_0} \big]$ produces an error of order $\co\big(n^{-s/2}\big)$ in the upper part of eq.together-indicator. In the lower part, the factor $n-u(k_\cdot)$ is missing and the order therefore even smaller.

Repeating this argument consecutively (note that this requires $c(r)$ steps at most) until no $\E_{\delta_n,n}\big[ \big(X_1^2-1 \big)^{l} \big]$ is left anymore, we arrive at

\begin{align*}%\label{eq.together-indicator}
%	\begin{split}
&n^{-(\alpha+r)} \sum_{*(k_\cdot,r,\cdot)} {n \choose k_{1},\ldots ,k_{r},(n-u(k_\cdot))} \frac{r!}{1!^{k_1}\cdots r!^{k_r}} \big(n-u(k_\cdot)\big) \1_{\{k_{\lfloor s/2\rfloor+1}=\dots=k_r=0\}}\\*
&\quad\qquad \cdot \bigg( \prod_{l=1}^{r}\Big(\E\Big[ \big(X_1^2-1 \big)^{l} \Big]\Big)^{k_l} \bigg) \E_{\delta_n,n}\Big[ |X_1|^{2\alpha} \Big] \P\big(\{|X_1|<\delta_n \sqrt{n} \, \}\big)^{n-u(k_\cdot)-1}\\
&\quad \quad + n^{-(\alpha+r)} \sum_{*(k_\cdot,r,\cdot)} {n \choose k_{1},\ldots ,k_{r},(n-u(k_\cdot))} \frac{r!}{1!^{k_1}\cdots r!^{k_r}} \\
&\quad \qquad \cdot \Bigg( \sum_{j=1}^{u(k_\cdot)} \1_{\{k_{\lfloor s/2\rfloor+1}=\dots=k_r=0,\, 2\alpha+2l^*\le{s}\}} \bigg( \prod_{l=1 , ~ l \ne l^*}^{r}\Big(\E\Big[ \big(X_1^2-1 \big)^{l} \Big]\Big)^{k_l} \bigg) \\
&\quad \qquad \cdot \Big(\E\Big[ \big(X_1^2-1 \big)^{l^*} \Big]\Big)^{k_{l^*}-1} \E_{\delta_n,n}\Big[ \big(X_1^2-1 \big)^{l^*}|X_1|^{2\alpha} \Big] \P\big(\{|X_1|<\delta_n \sqrt{n} \, \}\big)^{n-u(k_\cdot)} \Bigg)\\*
&\quad \quad +\co \big(n^{-({s}-2)/2}\big).
%	\end{split}
\end{align*}

Similar in spirit, we consecutively transition from $\P\big(\{|X_1|\ge\delta_n \sqrt{n} \, \}\big)$ to 1 at the cost of at most $\co \big(n^{-{s}/2}\big)$, from $\E_{\delta_n,n}\Big[ |X_1|^{2\alpha} \Big]$ to $\E\Big[ |X_1|^{2\alpha} \Big]$ at the cost of at most $\co \big(n^{-({s}-2)/2}\big)$ because

\begin{align*}
-\alpha-r+u(k_\cdot)+1-s/2+\alpha \le -(s-2)/2,
\end{align*}

and from $\E_{\delta_n,n}\Big[ \big(X_1^2-1 \big)^{l^*}|X_1|^{2\alpha} \Big]$ to $\E\Big[ \big(X_1^2-1 \big)^{l^*}|X_1|^{2\alpha} \Big]$ at the cost of at most $\co \big(n^{-({s}-2)/2}\big)$ because

\begin{align*}
&\le -s/2-r+(k_1+\dots+k_{l^*}+\dots+k_r)+({l^*}-1)k_{l^*}+1\\
&= -s/2-r+(k_1+\dots+{l^*}k_{l^*}+\dots+k_r)+1\\
&\le -(s-2)/2.
\end{align*}

As a consequence, we get from eq.together-indicator

\begin{align}\label{eq.together}
%	\begin{split}
\lefteqn{n^{-(\alpha+r)} \E\Big[ \1_{A_{n,1}} \Big(\sum (X_j^2-1)\Big)^{r}\Big(\sum|X_j|^{2\alpha}\Big)\Big]}\quad\nonumber\\*
&=n^{-(\alpha+r)} \sum_{*(k_\cdot,r,\cdot)} {n \choose k_{1},\ldots ,k_{r},(n-u(k_\cdot))} \frac{r!}{1!^{k_1}\cdots r!^{k_r}} \nonumber\\
&\qquad \cdot \big(n-u(k_\cdot)\big) \1_{\{k_{\lfloor s/2\rfloor+1}=\dots=k_r=0\}} \bigg( \prod_{l=1}^{r}\Big(\E\Big[ \big(X_1^2-1 \big)^{l} \Big]\Big)^{k_l} \bigg) \E\Big[ |X_1|^{2\alpha} \Big]\nonumber\\
&\quad+ n^{-(\alpha+r)} \sum_{*(k_\cdot,r,\cdot)} {n \choose k_{1},\ldots ,k_{r},(n-u(k_\cdot))} \frac{r!}{1!^{k_1}\cdots r!^{k_r}} \\
& \qquad \cdot \sum_{j=1}^{u(k_\cdot)} \1_{\{k_{\lfloor s/2\rfloor+1}=\dots=k_r=0,\, 2\alpha+2l^*\le{s}\}} \bigg( \prod_{l=1 , ~ l \ne l^*}^{r}\Big(\E\Big[ \big(X_1^2-1 \big)^{l} \Big]\Big)^{k_l} \bigg) \nonumber\\
& \qquad \cdot \Big(\E\Big[ \big(X_1^2-1 \big)^{l^*} \Big]\Big)^{k_{l^*}-1} \E\Big[ \big(X_1^2-1 \big)^{l^*}|X_1|^{2\alpha} \Big]\nonumber\\*
& \quad +\co \big(n^{-({s}-2)/2}\big).\nonumber
%	\end{split}
\end{align}

Now we return to the remainder of the Taylor expansion in eq.E-expanded and shall prove that this term is of order $\co\big(n^{-({s}-2)/2}\big)$. Recall that $\xi$ is a (random) intermediate value between $1$ and $n^{-1} V_n^2$ and ${h}=2\lceil \frac{m-2\alpha+1}{2}\rceil$. Note that $\xi\ge 1/2$ on $A_{n,2}$ and since $2\le\alpha\le m/2$, $\1_{A_{n,2}}\xi^{-(\alpha+{h})}\le c(m)$. Together with $|t_{{h},2\alpha}|\le c(m)$, this yields

\begin{align*}
%	\begin{split}
&\bigg|\E\Big[ \1_{A_{n,1}\cap A_{n,2}} (-1)^{{h}} n^{-(\alpha+{h})} t_{{h},2\alpha} \xi^{-(\alpha+{h})} \Big(\sum (X_j^2-1)\Big)^{{h}}\Big(\sum|X_j|^{2\alpha}\Big) \Big]\bigg| \\
&\quad\le c(m) n^{-(\alpha+{h})} \, \E\Big[ \1_{A_{n,1}} \Big(\sum (X_j^2-1)\Big)^{{h}}\Big(\sum|X_j|^{2\alpha}\Big) \Big].
%	\end{split}
\end{align*}

If we trace through the arguments above, we see that they are valid for all $r=0,\dots,m$. Thus, all steps are also valid for the remainder of the Taylor expansion, where $r={h}$. By eq.together, the expression above thus is equal to

\begin{align}\label{eq.remainder-together}
%	\begin{split}
&c(m)\Bigg(n^{-(\alpha+r)} \sum_{*(k_\cdot,r,\cdot)} {n \choose k_{1},\ldots ,k_{r},(n-u(k_\cdot))} \frac{r!}{1!^{k_1}\cdots r!^{k_r}} \nonumber\\*
&\qquad \cdot \big(n-u(k_\cdot)\big) \1_{\{k_{\lfloor s/2\rfloor+1}=\dots=k_r=0\}} \bigg( \prod_{l=1}^{r}\Big(\E\Big[ \big(X_1^2-1 \big)^{l} \Big]\Big)^{k_l} \bigg) \E\Big[ |X_1|^{2\alpha} \Big]\nonumber\\
&\quad + n^{-(\alpha+r)} \sum_{*(k_\cdot,r,\cdot)} {n \choose k_{1},\ldots ,k_{r},(n-u(k_\cdot))} \frac{r!}{1!^{k_1}\cdots r!^{k_r}} \\
&\qquad \cdot \sum_{j=1}^{u(k_\cdot)} \1_{\{k_{\lfloor s/2\rfloor+1}=\dots=k_r=0,\, 2\alpha+2l^*\le{s}\}} \bigg( \prod_{l=1 , ~ l \ne l^*}^{r}\Big(\E\Big[ \big(X_1^2-1 \big)^{l} \Big]\Big)^{k_l} \bigg) \nonumber\\
&\qquad \cdot \Big(\E\Big[ \big(X_1^2-1 \big)^{l^*} \Big]\Big)^{k_{l^*}-1} \E\Big[ \big(X_1^2-1 \big)^{l^*}|X_1|^{2\alpha} \Big]\Bigg)\nonumber\\*
&\quad +\co \big(n^{-({s}-2)/2}\big)\nonumber
%	\end{split}
\end{align}

with $r={h}$.
Note that $\E X_1^2=1$. Thus in the first part, all summands with $k_1>0$ vanish and in the second part, all summands with $k_1>1$ vanish. This results in

\[
u(k_\cdot)=k_1+k_2+\dots+k_r=\tfrac12(2k_2+\dots+2k_r)\le r/2
\]

for the first part and

\[
u(k_\cdot)=k_1+k_2+\dots+k_r\le1+\tfrac12(2k_2+\dots+2k_r)\le 1+r/2
\]

for the second part.

Now we want to determine the order in $n$ of the terms in eq.remainder-together. Note that all appearing moments are finite. Using $r={h}\ge m-2\alpha+1$, we can focus on the power of $n$ which is in the first part

\begin{align*}
&\le -\alpha-r+r/2+1
\end{align*}

and in the second part

\begin{align*}
\le -\alpha-r+1+r/2
\le-(m-1)/2.
\end{align*}

Thus, all terms (at most $c(m)$ many) from eq.remainder-together are of order $\co \big(n^{-({s}-2)/2}\big)$.

Ultimately, merging eq.E-expanded,eq.A_n,2-raus,eq.together,eq.remainder-together, we arrive at

\begin{align}\label{eq.all}
%	\begin{split}
\lefteqn{\E\Big[ V_n^{-2\alpha}\sum|X_j|^{2\alpha} \Big]}\quad\nonumber\\
&=\sum_{r=0}^{{h}-1} (-1)^r n^{-(\alpha+r)} t_{r,2\alpha} \sum_{*(k_\cdot,r,\cdot)} {n \choose k_{1},\ldots ,k_{r},(n-u(k_\cdot))} \frac{r!}{1!^{k_1}\cdots r!^{k_r}} \nonumber\\
&\qquad \cdot \big(n-u(k_\cdot)\big) \1_{\{k_{\lfloor s/2\rfloor+1}=\dots=k_r=0\}} \bigg( \prod_{l=1}^{r}\Big(\E\Big[ \big(X_1^2-1 \big)^{l} \Big]\Big)^{k_l} \bigg) \E\Big[ |X_1|^{2\alpha} \Big]\nonumber\\
&\quad+ \sum_{r=0}^{{h}-1} (-1)^r n^{-(\alpha+r)} t_{r,2\alpha} \sum_{*(k_\cdot,r,\cdot)} {n \choose k_{1},\ldots ,k_{r},(n-u(k_\cdot))} \frac{r!}{1!^{k_1}\cdots r!^{k_r}} \\
&\qquad \cdot \sum_{j=1}^{u(k_\cdot)} \1_{\{k_{\lfloor s/2\rfloor+1}=\dots=k_r=0,\, 2\alpha+2l^*\le{s}\}} \bigg( \prod_{l=1 , ~ l \ne l^*}^{r}\Big(\E\Big[ \big(X_1^2-1 \big)^{l} \Big]\Big)^{k_l} \bigg) \nonumber\\
&\qquad \cdot \Big(\E\Big[ \big(X_1^2-1 \big)^{l^*} \Big]\Big)^{k_{l^*}-1} \E\Big[ \big(X_1^2-1 \big)^{l^*}|X_1|^{2\alpha} \Big]\nonumber\\
&\quad +\co \big(n^{-({s}-2)/2}\big).\nonumber
%	\end{split}
\end{align}

All expectations in eq.all can be calculated explicitly as no power of $X_1$ is greater than $s$. Using $u(k_\cdot)\le r, \alpha\in\N$ and $\alpha\ge 2$, this results in an expansion in terms of orders $n^{-1},n^{-2},\dots,n^{-\lfloor m-2/2\rfloor}$ with moments up to order $m$ and a remainder of order $\co \big(n^{-({s}-2)/2}\big)$. By eq.tk-2r,eq.tlambda for $k\ge4$, the expectation of $\tlambda_{k,n}$ and its powers admit the same expansion up to the respective coefficient from eq.tk-2r. In <ref>, this expansion is performed exemplarily for $k=4$ and $k=6$. By eq.Phi^tP and eq.tP, $\E\big[\Phi^{\tP}_{m,n}(x)\big]$ admits a similar representation. By the definition of $Q_r$ in eq.Q-def, the expansion terms of order $n^{-r/2}$ are assigned to their respective $Q_r$ (or to the remainder $\co \big(n^{-({s}-2)/2}\big)$). Note that $Q_{r}=0$ for uneven $r$. That is, all the non-remainder terms in
are exactly the summands appearing in $\Phi^Q_{m,n}$, defined in eq.Phi^Q-def, and therefore cancel each other out in eq.E-tP-Q,eq.E-tp-q.
What still needs to be examined is the dependence of the remainder on the argument $x$. In $\tP_{k,n}$ (as well as in $\tp_{k,n}$), the $\tlambda_{k,n}$ are multiplied by $\phi$ and a Hermite polynomial of order $l\le2m$. For $l\le2m$, we can bound

\[
\sup_{x} \exp(x^2/4)\phi(x)H_l(x)\le c(l).
\]

As this non-random factor remains unaffected by the expectation, it also appears in the remainder such that eq.E-tP-Q,eq.E-tp-q hold.

§.§ Further proofs for Section <ref>

We prove this lemma by an approach similar to the one used in the classical proof of Markov's inequality. Without loss of generality let $a_1=\max\{a_1,\dots,a_l\}$. By using $M_{n,l}=\max_{j= l+1,l+2,\dots,n} |X_j|$ and $j_0\in\{1,\dots,n\}$ being the smallest index with $|X_{j_0}|=M_n$, we see

\begin{align*}
\lefteqn{ \E\Big[\1_{\big\{M_n\ge \sqrt{\frac{n}{\log(n)\, \eta}}\big\}} |X_1|^{a_1} \cdots |X_l|^{a_l}\Big]}\quad\\
%&= \E\Big[\sum \1_{\{j_0=j\}}\1_{\big\{M_n\ge \sqrt{\frac{n}{\log(n)\, \eta}}\big\}} |X_1|^{a_1} \cdots |X_l|^{a_l}\Big]\\
&= \sum_{j=1}^l \E\Big[ \1_{\{j_0=j\}}\1_{\big\{M_n\ge \sqrt{\frac{n}{\log(n)\, \eta}}\big\}} |X_1|^{a_1} \cdots |X_l|^{a_l}\Big]\\
&\quad+ \E\Big[ \sum_{j=l+1}^n \1_{\{j_0=j\}}\1_{\big\{M_n\ge \sqrt{\frac{n}{\log(n)\, \eta}}\big\}} |X_1|^{a_1} \cdots |X_l|^{a_l}\Big]\\
&\le l \, \E\Big[\1_{\{j_0=1\}}\1_{\big\{M_n\ge \sqrt{\frac{n}{\log(n)\, \eta}}\big\}} |X_1|^{a_1} \cdots |X_l|^{a_l}\Big]\\
&\quad+ \E\Big[\1_{\{j_0>l\}}\1_{\big\{M_n\ge \sqrt{\frac{n}{\log(n)\, \eta}}\big\}} |X_1|^{a_1} \cdots |X_l|^{a_l}\Big]\\
&\le \, l \, \big(\tfrac{n}{\log(n)\, \eta}\big)^{-({s}-a_1)/2} \, \E\Big[\1_{\{j_0=1\}}\1_{\big\{M_n\ge \sqrt{\frac{n}{\log(n)\, \eta}}\big\}} |X_1|^{{s}} |X_2|^{a_2} \cdots |X_l|^{a_l}\Big]\\
&\quad+ \E\Big[\1_{\{j_0>l\}}\1_{\big\{M_{n,l}\ge \sqrt{\frac{n}{\log(n)\, \eta}}\big\}} |X_1|^{a_1} \cdots |X_l|^{a_l}\Big]\\
&\le \, l \, \big(\tfrac{n}{\log(n)\, \eta}\big)^{-({s}-a_1)/2} \, \E\Big[ |X_1|^{{s}} |X_2|^{a_2} \cdots |X_l|^{a_l}\Big]\\
&\quad+ \E\Big[\1_{\big\{M_{n,l}\ge \sqrt{\frac{n}{\log(n)\, \eta}}\big\}} |X_1|^{a_1} \cdots |X_l|^{a_l}\Big]\\
&= l \, \eta^{({s}-a_1)/2}\, (\log n)^{({s}-a_1)/2} n^{-({s}-a_1)/2} \E |X_1|^{{s}} \E |X_1|^{a_2} \cdots \E |X_1|^{a_l}\\
&\quad+ \P\big(M_{n,l}\ge \sqrt{\tfrac{n}{\log(n)\, \eta}} \, \big) \E |X_1|^{a_1} \cdots \E |X_1|^{a_l}\\
%	&\le (\log n)^{{s}-a_1} n^{-({s}-a_1)/2} l \E |X_1|^{{s}} \E |X_1|^{a_2} \cdots \E |X_1|^{a_l}\\
%	&\quad+ (n-l)\P\big(|X_1|\ge \sqrt{\tfrac{n}{\log(n)\, \eta}} \, \big) \E |X_1|^{a_1} \cdots \E |X_1|^{a_l}\\
%	&\le (\log n)^{{s}-a_1} n^{-({s}-a_1)/2} l \E |X_1|^{{s}} \E |X_1|^{a_2} \cdots \E |X_1|^{a_l}\\
%	&\quad+ (n-l)\E |X_1|^{{s}}n^{-{s}/2}(\log n)^{{s}} \E |X_1|^{a_1} \cdots \E |X_1|^{a_l}\\
%	&= \cO\Big( n^{-({s}-a_1)/2} \, (\log n)^{{s}-a_1} \Big)\\
%	&\quad+ \co\big( n^{-({s}-2)/2} \, (\log n)^{{s}} \big)\\
& = \co\Big( n^{-({s}-\max\{2,a_1\})/2} \, (\log n)^{{s}/2} \Big),
\end{align*}

where we used eq.Mn-an in the last step.

Using $(\delta_n)$ from <ref>, we get by eq.Lle1, eq.Vn-bound-1/2 and eq.Mn-delta

\begin{align*}
\E\left[ \tL_{k,n}\right]
%&=\E\left[\1_{\{V_n^2 \le \frac n2\}} \tL_{k,n}\right]\\
%&\quad+\E\left[ \1_{\{V_n^2 > n/2, M_n\ge \delta_n \sqrt{n} \, \}} \tL_{k,n} \right]\\
%&\quad+\E\left[ \1_{\{V_n^2 > n/2, M_n< \delta_n \sqrt{n} \, \}} \tL_{k,n} \right]\\
&\le\E\left[ \1_{\{V_n^2 \le \frac n2\}}\right]\\
&\quad+\E\left[ \1_{\{M_n\ge \delta_n \sqrt{n} \, \}}\right]\\
&\quad+\E\left[ \1_{\{V_n^2 > n/2, M_n< \delta_n \sqrt{n} \, \}} V_n^{-k}M_n^{k-{s}}\sum|X_j|^{{s}} \right]\\
%&\le\P\left(V_n^2 \le \tfrac n2\right)\\*
%&\quad+\P\left(M_n\ge \delta_n \sqrt{n} \, \right)\\
%&\quad+\E\left[ \1_{\{V_n^2 > n/2, M_n< \delta_n \sqrt{n} \, \}} \big(\tfrac n2\big)^{-k/2}\big(\delta_n \sqrt{n}\big)^{k-{s}}\sum|X_j|^{{s}}\right]\\
&\quad+ \co\big(n^{-({s}-2)/2}\big)\\
&\quad+ 2^{k/2} n^{-({s}-2)/2} \delta_n^{k-{s}} \E\big[ |X_1|^{{s}}\big]\\
&= \co\big(n^{-({s}-2)/2}\big)
\end{align*}

since $\delta_n\to0$.

We split up the expectation and get by eq.Vn-bound-1/2,eq.Mn-an

\begin{align*}
\E\left[ \exp\big(-\eta\,B_n^2\big)\right]
%&=\E\left[ \1_{\{V_n^2 \le \frac n2\}} \exp\big(-\eta\,B_n^2\big) \right]\\
%&\quad+\E\left[ \1_{\big\{V_n^2 > n/2, M_n\ge \sqrt{\frac{n\,\eta}{\log(n)\, {s}}}\big\}} \exp\big(-\eta\,B_n^2\big) \right]\\
%&\quad+\E\left[ \1_{\big\{V_n^2 > n/2, M_n< \sqrt{\frac{n\,\eta}{\log(n)\, {s}}}\big\}} \exp\big(-\eta\,B_n^2\big) \right]\\
&\le\E\left[ \1_{\{V_n^2 \le \frac n2\}}\right]\\
&\quad+\E\left[ \1_{\big\{M_n\ge \sqrt{\frac{n\,\eta}{\log(n)\, {s}}}\big\}}\right]\\
&\quad+\E\left[ \1_{\big\{V_n^2 > n/2, M_n< \sqrt{\frac{n\,\eta}{\log(n)\, {s}}}\big\}} \exp\big(-\eta\,V_n^2\,M_n^{-2}\big) \right]\\
&\le\P\left(V_n^2 \le \tfrac n2\right)\\
&\quad+\P\left(M_n\ge \sqrt{\tfrac{n\,\eta}{\log(n)\, {s}}}\,\right)\\
&\quad+\E\left[ \1_{\big\{V_n^2 > n/2, M_n< \sqrt{\frac{n\,\eta}{\log(n)\, {s}}}\big\}} \exp\big(-\eta\,\tfrac n2\,\tfrac{\log(n)\,s}{n\,\eta}\big)\right]\\
&\quad+\co\big( n^{-({s}-2)/2} \, (\log n)^{{s}/2} (\eta\wedge 1)^{-s/2} \big)\\
&= \co\big( n^{-({s}-2)/2} \, (\log n)^{{s}/2} (\eta\wedge 1)^{-s/2} \big).
\end{align*}

§.§ Further proofs for Section <ref>

Take $f$ to be the density of $X_1$, $Y_j:=(X_j,X_j^2)$, $B:=B_1\times B_2:=[a_1,b_1]\times[a_2,b_2]$ with $a_1<b_1$, $a_2<b_2$ in $\R$. Then by substitution and Fubini's theorem,

\begin{align*}
\lefteqn{\bP\big(Y_1+Y_2+Y_3 \in B\big)}\quad\\*
&=\bP\Big(X_1+X_2+X_3 \in B_1, X_1^2+X_2^2+X_3^2 \in B_2\Big)\\
&=2\int_\R\int_\R\int_\R f(x)f(y)f(z)\1\big\{x+y+z\in B_1, x^2+y^2+z^2\in B_2, x\le y\big\}\dx\dy\dz\\
&=2\int_\R\int_\R\int_\R f(s-y-z)f(y)f(z)\\*
&\qquad\qquad\quad\cdot\1\big\{s\in B_1, (s-y-z)^2+y^2+z^2 \in B_2, (s-z)/2 \le y\big\}\ds\dy\dz\\
&=2\int_{B_1}\int_\R\int_\R f(s-y-z)f(y)f(z)\\*
&\qquad\qquad\quad\cdot\1\big\{ (s-y-z)^2+y^2+z^2 \in B_2, (s-z)/2 \le y\big\}\dy\dz\ds.
\end{align*}

\[
v_{s,z}(y):=(s-y-z)^2+y^2+z^2=\tfrac12 \Big( \big(2y-(s-z)\big)^2+s^2+3z^2-2sz\Big).
\]

Note that

\[
v'_{s,z}(y) = 2 (2y-s+z).
\]

As we only consider the regime $(s-z)/2 \le y$, the inverse is

\[
v_{s,z}^{-1}(v)=\tfrac12 \Big(s-z+\sqrt{2v-s^2-3z^2+2sz}\Big)=\tfrac12 \Big(s-z+\tfrac{1}{\sqrt{3}} \sqrt{6v-2s^2-(3z-s)^2}\Big).
\]

Thus the expression above is equal to

\begin{align*}
&2\int_{B_1}\int_\R\int_{(s-z)/2}^\infty \frac{2(2v_{s,z}^{-1}(v_{s,z}(y))-s+z)}{2(2v_{s,z}^{-1}(v_{s,z}(y))-s+z)}f(s-v_{s,z}^{-1}(v_{s,z}(y))-z)f(v_{s,z}^{-1}(v_{s,z}(y)))f(z)\\*
&\qquad\qquad\qquad\quad\cdot\1\big\{ v_{s,z}(y) \in B_2 \big\}\dy\dz\ds\\*
&=2\int_{B_1}\int_\R\int_{v_{s,z}((s-z)/2)}^\infty \frac{1}{2(2v_{s,z}^{-1}(v)-s+z)} f\big(s-v_{s,z}^{-1}(v)-z\big) f(v_{s,z}^{-1}(v)) f(z) \\*
&\qquad\qquad\qquad\qquad\qquad\cdot\1\big\{ v \in B_2\big\}\dv\dz\ds\\
&=\int_{B_1}\int_{B_2}\int_\R \frac{1}{2v_{s,z}^{-1}(v)-s+z} f\big(s-v_{s,z}^{-1}(v)-z\big) f(v_{s,z}^{-1}(v)) f(z)\\*
&\qquad\qquad\qquad\cdot\1\big\{ v_{s,z}\big((s-z)/2\big) \le v \big\}\dz\dv\ds.
\end{align*}

Hence the random variable $(Y_1+Y_2+Y_3)$ has the density

\[
\int_\R \frac{1}{2v_{s,z}^{-1}(v)-s+z} f\big(s-v_{s,z}^{-1}(v)-z\big) f(v_{s,z}^{-1}(v)) f(z) \1\big\{ v_{s,z}\big((s-z)/2\big) \le v \big\}\dz
\]

for $s,v\in\R$.
Note that

\[
v_{s,z}((s-z)/2) \le v \quad \Leftrightarrow \quad 6v-2s^2-(3z-s)^2 \ge 0 \quad \Rightarrow \quad 6v-2s^2 \ge 0.
\]

We examine the integral

\begin{align*}
\lefteqn{\int_\R \frac{1}{2v_{s,z}^{-1}(v)-s+z} \1\big\{ v_{s,z}((s-z)/2) \le v \big\}\dz}\quad\\*
&=\int_\R \frac{\sqrt{3}}{\sqrt{6v-2s^2-(3z-s)^2}} \1\big\{ \tfrac12 \big( s^2+3z^2-2sz\big) \le v \big\}\dz\\
&=\int_\R \frac{\sqrt{3}}{\sqrt{6v-2s^2-(3z-s)^2}} \1\big\{ \tfrac13 \big(s-\sqrt{6v-2s^2} \big) \le z \le \tfrac13 \big(s+\sqrt{6v-2s^2}\big) \big\}\dz\\
&=\int_{\frac13 (s-\sqrt{6v-2s^2})}^{\frac13 (s+\sqrt{6v-2s^2})} \frac{\sqrt{3}}{\sqrt{6v-2s^2-(3z-s)^2}}\dz\\
&=\frac{1}{\sqrt{3}}\int_{-\sqrt{6v-2s^2}}^{\sqrt{6v-2s^2}} \frac{1}{\sqrt{6v-2s^2-w^2}}\dw=\frac{\pi}{\sqrt{3}}
\end{align*}

by <ref>.
Therefore for all bounded $f$, $(\sum X_j,\sum X_j^2)$ has a bounded density for $n=3$ (or equivalently, for all $n\ge3$). Thus by <cit.>, <ref> is fulfilled.

By the triangle inequality and <ref>, we get for ${k=0,\dots,m}$

\begin{align*}
%	\begin{split}
\lefteqn{\Big|\ddtk \Big(\tphi_{T_n'}(t)-\tphi_{\Phi^{\tP}_{m,n}}(t)\Big)\Big|}\quad\\
%&\le \Big|\ddtk \Big(\tphi_{T_n'}(t)-\tphi_{T_n}(t)\Big)\Big| +\Big|\ddtk \Big(\tphi_{T_n}(t)-\tphi_{\Phi^{\tP}_{m,n}}(t)\Big)\Big|\\
&\le \Big|\ddtk \Big(\tphi_{T_n'}(t)-\tphi_{T_n}(t)\Big)\Big|+ c(m)\tL_{m+1,n}e^{-t^2/6}\big(|t|^{m+1-k}+|t|^{3m-1+k}\big),
%	\end{split}
\end{align*}

where we expand the first summand by the Leibniz rule as

\begin{align*}
\Big|\ddtk \Big(\tphi_{T_n'}(t)-\tphi_{T_n}(t)\Big)\Big|
%&= \Big| \sum_{l=0}^k \binom{k}{l}\Big(\ddtl\tphi_{T_n}(t)\Big)\Big(\ddT{k-l}\exp(-\tfrac12 \beta_n t^2)\Big) -\Big(\ddtk\tphi_{T_n}(t)\Big)\Big|\\
%	&= \Big| \sum_{l=0}^{k-1} \binom{k}{l}\Big(\ddtl\tphi_{T_n}(t)\Big)\Big(\ddT{k-l}\exp(-\tfrac12 \beta_n t^2)\Big)\\
%	&\quad+\Big(\ddtk\tphi_{T_n}(t)\Big)\Big(\exp(-\tfrac12 \beta_n t^2)-1\Big)\Big|\\
&\le \sum_{l=0}^{k-1} \binom{k}{l}\Big|\ddtl\tphi_{T_n}(t)\Big|\Big|\ddT{k-l}\exp(-\tfrac12 \beta_n t^2)\Big|\\
&\quad+\Big|\ddtk\tphi_{T_n}(t)\Big|\Big(1-\exp(-\tfrac12 \beta_n t^2)\Big).
\end{align*}

By <ref>, eq.ddtl-tphi and eq.Lle1,

\begin{align*}
\Big|\ddtl\tphi_{T_n}(t)\Big|
&\le \Big|\ddtl\tphi_{\Phi^{\tP}_{m,n}}(t)\Big|+c(m)\tL_{m+1,n}e^{-t^2/6}\big(|t|^{m+1-l}+|t|^{3m-1+l}\big)\\
&\le c(m)e^{-t^2/2}\big(1+|t|^{2m-4+l}\big)+c(m)e^{-t^2/6}\big(|t|^{m+1-l}+|t|^{3m-1+l}\big)\\
&\le c(m)e^{-t^2/6}\big(1+|t|^{3m-1+l}\big)
\end{align*}

for $l=0,\dots,m$. Additionally for $k>l$,

\[\Big|\ddT{k-l}\exp(-\tfrac12 \beta_n t^2)\Big|
\le c(k-l) \beta_n (1+|t|^{k-l}) \exp(-\tfrac12 \beta_n t^2)
\]

\[\Big(1-\exp(-\tfrac12 \beta_n t^2)\Big)\le\tfrac12 \beta_n t^2.\]

Putting everything together yields

\begin{align*}
\lefteqn{\Big|\ddtk \Big(\tphi_{T_n'}(t)-\tphi_{T_n}(t)\Big)\Big|}\quad\\*
&\le \sum_{l=0}^{k-1} \binom{k}{l}\Big|\ddtl\tphi_{T_n}(t)\Big|\Big|\ddT{k-l}\exp(-\tfrac12 \beta_n t^2)\Big|\\
&\quad+\Big|\ddtk\tphi_{T_n}(t)\Big|\Big(1-\exp(-\tfrac12 \beta_n t^2)\Big)\\
&\le \sum_{l=0}^{k-1} \binom{k}{l}\Big(c(m)e^{-t^2/6}\big(1+|t|^{3m-1+l}\big)\Big)\Big(c(k-l) \beta_n (1+|t|^{k-l}) \exp(-\tfrac12 \beta_n t^2)\Big)\\
&\quad+\Big(c(m)e^{-t^2/6}\big(1+|t|^{3m-1+k}\big)\Big)\Big(\tfrac12 \beta_n t^2\Big)\\
&\le c(m) \beta_n e^{-t^2/6} \big(1+|t|^{3m+1+k}\big),
\end{align*}

which proves eq.lemma4'.

§.§ Further remarks

Let $\Sigma$ be the covariance matrix of $(X_1,X_1^2)$ and assume $\mu_4<\infty$. Then the following equivalences hold:

$(X_1,X_1^2)$ is non-degenerate

$\Leftrightarrow$ $\Var\big(a_1 X_1+a_2 X_1^2)>0$ for all $a=(a_1,a_2)\in\R^2$ with $a\ne0$

$\Leftrightarrow$ $a^T \Sigma a >0$ for all $a\in\R^2$ with $a\ne0$

$\Leftrightarrow$ $\Sigma$ is positive definite.

In this remark, we do neither assume that $X_1$ is symmetric, nor that $\mu_2=1$. Under these conditions, we investigate if $T_n$ is normalized, expanding <ref>. For clear illustration however, let all moments be finite in the following procedure. By eq.V_n^-2 we expand

\begin{equation}\label{eq.Tn-exp}
\begin{split}
%&=n^{-1}S_n^2\Big(\mu_2+n^{-1} \Big(\sum (X_j^2-\mu_2)\Big)\Big)^{-1}\\
&=n^{-1} \mu_2^{-1} S_n^2
- n^{-2} \mu_2^{-2} S_n^2 \Big(\sum (X_j^2-\mu_2)\Big)\\
&\quad+ n^{-3} \mu_2^{-3} S_n^2 \Big(\sum (X_j^2-\mu_2)\Big)^2
+ \cO_p\big(n^{-2}\big).
\end{split}
\end{equation}

Due to $\mu_1=0$ and $\E X_j^2=\mu_2$, the index of every factor within

\[
\Big(\sum X_j\Big)^{2}\Big(\sum (X_j^2-\mu_2)\Big)^{k}
=\sum_{j_1,\dots,j_{k+2}=1}^n X_{j_1} X_{j_2} \big(X_{j_3}^2-\mu_2\big) \cdots \big( X_{j_{k+2}}^2-\mu_2 \big)
\]

has to be equal to the index of another factor or the summand vanishes within the expectation. We take the expectation of each summand of eq.Tn-exp separately

\begin{align*}
\E S_n^2
&= n \mu_2,\\
\E \Big[ S_n^2 \Big(\sum (X_j^2-\mu_2)\Big) \Big]
&= n (\mu_4-\mu_2^2),\\
\E \Big[ S_n^2 \Big(\sum (X_j^2-\mu_2)\Big)^2 \Big]
%&= n^2 \big(\mu_2(\mu_4-2\mu_2^2+\mu_2^2)+2(\mu_3-\mu_1 \mu_2)^2\big)
%+ \cO(n)\\
&= n^2 \big(\mu_2\mu_4-\mu_2^3+2\mu_3^2\big)
+ \cO(n)
\end{align*}

which leads to

\begin{align*}
\E T_n^2
%	&=n^{-1} \mu_2^{-1} n \mu_2\\
%	&- n^{-2} \mu_2^{-2} n (\mu_4-\mu_2^2)\\
%	&+ n^{-3} \mu_2^{-3} \Big(n^2 \big(\mu_2\mu_4-\mu_2^3+2\mu_3^2\big) + \cO(n)\Big)\\
%	&+ \cO\big(n^{-2}\big)\\
%	&= 1\\
%	&+ n^{-1} \mu_2^{-3} (-\mu_4\mu_2+\mu_2^3+\mu_2\mu_4-\mu_2^3+2\mu_3^2)\\
%	&+ \cO\big(n^{-2}\big)\\
&= 1 + n^{-1} 2 \mu_2^{-3} \mu_3^2 + \cO\big(n^{-2}\big).
\end{align*}

§.§ Auxiliary results

Let $\nu_n\ge1$, $k=0,1,\dots$ and $0<\beta<4$. Then

\begin{equation*}
\int_{\nu_n}^\infty t^k \exp\big(- \beta t^2 /2 \big)\dt
\le c(k)\beta^{-(k+2)/2} \exp\big(- \beta \nu_n^2 /4 \big).
\end{equation*}

We examine

\begin{align*}
\int_{\nu_n}^\infty t^k \exp(- \beta t^2 /2)\dt
&\le \int_{\nu_n}^\infty t \, (t^2)^{\lceil\frac{k-1}{2}\rceil} \exp(- \beta t^2 /2)\dt\\
&= \frac{\lceil\frac{k-1}{2}\rceil!}{(\beta/4 )^{\lceil\frac{k-1}{2}\rceil}} \int_{\nu_n}^\infty t \, \frac{(\beta t^2 /4)^{\lceil\frac{k-1}{2}\rceil}}{\lceil\frac{k-1}{2}\rceil!} \exp(- \beta t^2 /2)\dt\\
&\le \frac{k!}{(\beta/4 )^{k/2}} \int_{\nu_n}^\infty t \Big(\sum_{l=0}^\infty \frac{(\beta t^2 /4)^{l}}{l!} \Big) \exp(- \beta t^2 /2)\dt\\
&= \frac{k!\,2^k}{\beta^{k/2}} \int_{\nu_n}^\infty t \, \exp(\beta t^2 /4) \exp(- \beta t^2 /2)\dt\\
%&= \frac{k!\,2^{k+1}}{\beta^{(k+2)/2}} \int_{\nu_n}^\infty (\beta t /2 ) \exp(- \beta t^2 /4)\dt\\
&= \frac{k!\,2^{k+1}}{\beta^{(k+2)/2}} \exp( \beta \nu_n^2 /4).
\end{align*}

Let $f:\R\to\R$ be a measurable function with $0\le f(x)\le1$ for all $x\in\R$ and let $\mu$ be a probability measure on $\R$. Assume that $\mu(\{x:f(x)<1\})>0$. Then

\[
\int_\R f(x)\dmu < 1.
\]

For $n\in\N$ we define $A_n:=\{x:f(x)<1-1/n\}$, which are increasing for ${n\to \infty}$. By continuity from below,

\[
\lim_{n\to\infty}\mu(A_n)=\mu\big(\{x:f(x)<1\}\big)>0.
\]

Thus, there exists an $N\in\N$ such that $\mu(A_N)>0$ and therefore,

\begin{align*}
\int_\R f(x)\dmu
=\int_{A_N} f(x)\dmu + \int_{A_N^c} f(x)\dmu
<(1-\tfrac{1}{N}) \mu(A_N) + \mu(A_N^c)
=1-\tfrac{1}{N} \mu(A_N)<1.
\end{align*}

Let $n\ge2$. Then

\begin{align*}
\int_{\frac{1}{2}\sqrt{n}}^{2\sqrt{n}} \big(1+(z-n^{1/2})^2\big)^{-a/2} \dz
\le \begin{cases}
c(0) \sqrt{n} , &\text{ if } a=0,\\
c(0) \log n , &\text{ if } a=1,\\
c(0) , &\text{ if } a=2.
\end{cases}
\end{align*}

For $a=0$, the statement is trivial. Otherwise,

\begin{align*}
\int_{\frac{1}{2}\sqrt{n}}^{2\sqrt{n}} \big(1+(z-n^{1/2})^2\big)^{-a/2} \dz
=\int_{-\frac{1}{2}\sqrt{n}}^{\sqrt{n}} \big(1+z^2\big)^{-a/2} \dz
\le2\int_{0}^{\sqrt{n}} \big(1+z^2\big)^{-a/2} \dz.
\end{align*}

Now for $a=1$ and the inverse hyperbolic sine $\arsinh$,

\begin{align*}
\int_{0}^{\sqrt{n}} \big(1+z^2\big)^{-1/2} \dz
= \arsinh(\sqrt{n})
= \log(\sqrt{n} + \sqrt{1 + n})
\le c(0) \log(n),
\end{align*}

and for $a=2$

\begin{align*}
\int_{0}^{\sqrt{n}} \big(1+z^2\big)^{-1} \dz
\le \tfrac \pi 2.
\end{align*}

For $L(x)=x\log x$ with $x>0$,

\begin{equation*}
L(1+u+v)=L(1+u)+v+\theta_1 u v+\theta_2 v^2
\end{equation*}

for $|u|\le 1/4, |v|\le1/4$ and $|\theta_j|\le2$ depending on $u$ and $v$.

By Taylor expansion around $v=0$,

\begin{align*}
&=(1+u)\Big(\log(1+u+0) + v (1+u+0)^{-1} \\*
&\qquad\qquad+ \tfrac{v^2}{2} (-1) (1+u+\xi_1v)^{-2} \Big)\\
&=(1+u)\log(1+u) + v + v^2 (1+u) (-\tfrac12) (1+u+\xi_1v)^{-2}
\end{align*}

with $\xi_1\in[0,1]$ depending on $u$ and $v$ and by Taylor expansion around $u+v=0$,

\begin{align*}
&=v\Big(\log(1+0) + (u+v) (1+\xi_2(u+v))^{-1} \Big)\\*
&= (uv+v^2) (1+\xi_2(u+v))^{-1}
\end{align*}

with $\xi_2\in[0,1]$ depending on $u$ and $v$.

\begin{align*}
&\le (1-1/2)^{-1}
=2,\ \ \
\ge (1+1/2)^{-1}
= 2/3
\end{align*}

\begin{align*}
(1+u) \tfrac{1}{2} (1+u+\xi_1v)^{-2}
&\le (1+u) \tfrac{1}{2} (1+u-1/4)^{-2}
\le (3/4) \tfrac{1}{2} (1/2)^{-2}
= 3/2,\\*
(1+u) \tfrac{1}{2} (1+u+\xi_1v)^{-2}
&\ge (1+u) \tfrac{1}{2} (1+u+1/4)^{-2}
\ge (5/4) \tfrac{1}{2} (3/2)^{-2}
= 5/18.
\end{align*}

(1+\xi_2(u+v))^{-1}-(1+u) \tfrac{1}{2} (1+u+\xi_1v)^{-2}
\in [-5/6,31/18]$.

Let $a,b,l>0$. Then

\begin{align*}
\int_{-l}^l (ax^2+b)^{-3/2}\dx = 2\big(al^2+b\big)^{-1/2}\,b^{-1}\,l.
\end{align*}

\begin{align*}
\int_{-l}^l (ax^2+b)^{-3/2}\dx
&=2b^{-3/2} \int_{0}^l (\tfrac ab x^2+1)^{-3/2}\dx\\
&=2b^{-3/2}\sqrt{b/a} \int_{0}^{l\sqrt{a/b}} (x^2+1)^{-3/2}\dx\\
%&=2b^{-3/2}\sqrt{b/a} \, \frac{x}{\sqrt{x^2+1}} \, \bigg|_{x=0}^{l\sqrt{a/b}}\\
&=2b^{-3/2}\sqrt{b/a} \, \frac{l\sqrt{a/b}}{\sqrt{l^2a/b+1}} = 2\big(al^2+b\big)^{-1/2}\,b^{-1}\,l.
\end{align*}

Let $a > 0$. Then

\begin{align*}
\int_{-\sqrt{a}}^{\sqrt{a}} (a-w^2)^{-1/2} \dw
= \pi.
\end{align*}

\begin{align*}
\int_{-\sqrt{a}}^{\sqrt{a}} (a-w^2)^{-1/2}\dw
&=2a^{-1/2} \int_{0}^{\sqrt{a}} (1-w^2/a)^{-1/2}\dw\\
&=2a^{-1/2} a^{1/2} \int_{0}^{1} (1-w^2)^{-1/2}\dw\\
&= 2 \arcsin(1)=\pi.
\end{align*}

This work was supported by the DFG research units 1735 and 5381.

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