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$\newcommand\dee{\Delta_{\mathrm{TV}}}\newcommand{\vpi}{\varphi}$Let $f_n$ denote the pdf of \begin{equation*} S_n:=\sqrt{\frac3n} \sum_{i=1}^n X_i \end{equation*} and let $\vpi$ denote the standard normal pdf. Then \begin{equation*} \dee(D_n,N(0,1))=\int|f_n-\vpi|\le I_{n1}+I_{n2}+I_{n3}, \tag{1}\label{1} \end{equation*} where \begin{equation*} I_{n1}:=\int_{|x|\le\sqrt{3n}}|f_n(x)-\vpi(x)|\,dx, \end{equation*} \begin{equation*} I_{n2}:=\int_{|x|>\sqrt{3n}}f_n(x)\,dx, \end{equation*} \begin{equation*} I_{n3}:=\int_{|x|>\sqrt{3n}}\vpi(x)\,dx. \end{equation*}

We bound $I_{n1}$ using the asymptotic expansion in the central limit theorem given (say) by Theorem 7 (with $k=5$ and $l=1$) on p. 175 in the book by Petrov. Noting also Lemma 10 on p. 173 and the expressions for $Q_{kn}(x)$ on p. 138 of the same book, as well as the fact that $EX_i=EX_i^3=0$ for all $i$, we see that \begin{equation*} f_n(x)=\vpi(x)\Big(1+\frac{P_2(x)}n\Big)+O\Big(\frac1{n^{3/2}}\Big) \end{equation*} uniformly in all real $x$, where $P_2$ is a certain polynomial. It follows that $I_{n1}=O(1/n)$. Also, $I_{n2}=0$, since $|S_n|\le\sqrt{\frac3n}\,n=\sqrt{3n}$. Finally, it is easy to see that $I_{n3}=o(1/n)$.

Thus, by \eqref{1},
\begin{equation*} \dee(D_n,N(0,1))=O(1/n), \tag{2}\label{2} \end{equation*} as desired.

In fact, $P_2(x)=\frac1{20}(3-6x^2+x^4)$, and hence, slightly modifying the above reasoning, we see that \begin{equation*} \dee(D_n,N(0,1))\sim\frac cn, \end{equation*} where \begin{equation} c:=\int\vpi|P_2|= \frac{e^{-3/2-\sqrt{3/2}}}{5\sqrt\pi} \Big(e^{\sqrt{6}} \sqrt{9-3\sqrt{6}}+\sqrt{9+3\sqrt{6}}\,\Big) \\ =0.140030\ldots. \end{equation}

Here is the graph $\big\{\big(n,\frac nc\,\dee(D_n,N(0,1))\big)\colon n\in\{3,\dots,15\}\big\}$:

Following the lines of the proof, it easy to see that \eqref{2} will hold whenever, say, the $X_i$'s are iid with $EX_i=EX_i^3=0$ and light enough distribution tails (in your case, the $X_i$'s are iid symmetric random variables with no distribution tails).

$\newcommand\dee{\Delta_{\mathrm{TV}}}\newcommand{\vpi}{\varphi}$Let $f_n$ denote the pdf of \begin{equation*} S_n:=\sqrt{\frac3n} \sum_{i=1}^n X_i \end{equation*} and let $\vpi$ denote the standard normal pdf. Then \begin{equation*} \dee(D_n,N(0,1))=\int|f_n-\vpi|\le I_{n1}+I_{n2}+I_{n3}, \tag{1}\label{1} \end{equation*} where \begin{equation*} I_{n1}:=\int_{|x|\le\sqrt{3n}}|f_n(x)-\vpi(x)|\,dx, \end{equation*} \begin{equation*} I_{n2}:=\int_{|x|>\sqrt{3n}}f_n(x)\,dx, \end{equation*} \begin{equation*} I_{n3}:=\int_{|x|>\sqrt{3n}}\vpi(x)\,dx. \end{equation*}

We bound $I_{n1}$ using the asymptotic expansion in the central limit theorem given (say) by Theorem 7 (with $k=5$ and $l=1$) on p. 175 in the book by Petrov. Noting also Lemma 10 on p. 173 and the expressions for $Q_{kn}(x)$ on p. 138 of the same book, as well as the fact that $EX_i=EX_i^3=0$ for all $i$, we see that \begin{equation*} f_n(x)=\vpi(x)\Big(1+\frac{P_2(x)}n\Big)+O\Big(\frac1{n^{3/2}}\Big) \end{equation*} uniformly in all real $x$, where $P_2$ is a certain polynomial. It follows that $I_{n1}=O(1/n)$. Also, $I_{n2}=0$, since $|S_n|\le\sqrt{\frac3n}\,n=\sqrt{3n}$. Finally, it is easy to see that $I_{n3}=o(1/n)$.

Thus, by \eqref{1},
\begin{equation*} \dee(D_n,N(0,1))=O(1/n), \tag{2}\label{2} \end{equation*} as desired.

In fact, $P_2(x)=\frac1{20}(3-6x^2+x^4)$, and hence, slightly modifying the above reasoning, we see that \begin{equation*} \dee(D_n,N(0,1))\sim\frac cn, \end{equation*} where \begin{equation} c:=\int\vpi|P_2|= \frac{e^{-3/2-\sqrt{3/2}}}{5\sqrt\pi} \Big(e^{\sqrt{6}} \sqrt{9-3\sqrt{6}}+\sqrt{9+3\sqrt{6}}\,\Big) \\ =0.140030\ldots. \end{equation}

Here is the graph $\big\{\big(n,\frac nc\,\dee(D_n,N(0,1))\big)\colon n\in\{3,\dots,15\}\big\}$:

Following the lines of the proof, it easy to see that \eqref{2} will hold whenever, say, the $X_i$'s are iid with $EX_i=EX_i^3=0$ and light enough distribution tails (in your case, the $X_i$'s are iid symmetric random variables with no distribution tails), provided that there is some natural $k$ such that $S_k$ has an absolutely continuous pdf with integrable derivative (in your case, $k=2$ will do).

$\newcommand\dee{\Delta_{\mathrm{TV}}}\newcommand{\vpi}{\varphi}$Let $f_n$ denote the pdf of \begin{equation*} S_n:=\sqrt{\frac3n} \sum_{i=1}^n X_i \end{equation*} and let $\vpi$ denote the standard normal pdf. Then \begin{equation*} \dee(D_n,N(0,1))=\int|f_n-\vpi|\le I_{n1}+I_{n2}+I_{n3}, \tag{1}\label{1} \end{equation*} where \begin{equation*} I_{n1}:=\int_{|x|\le\sqrt{3n}}|f_n(x)-\vpi(x)|\,dx, \end{equation*} \begin{equation*} I_{n2}:=\int_{|x|>\sqrt{3n}}f_n(x)\,dx, \end{equation*} \begin{equation*} I_{n3}:=\int_{|x|>\sqrt{3n}}\vpi(x)\,dx. \end{equation*}

We bound $I_{n1}$ using the asymptotic expansion in the central limit theorem given (say) by Theorem 7 (with $k=5$ and $l=1$) on p. 175 in the book by Petrov. Noting also Lemma 10 on p. 173 and the expressions for $Q_{kn}(x)$ on p. 138 of the same book, as well as the fact that $EX_i=EX_i^3=0$ for all $i$, we see that \begin{equation*} f_n(x)=\vpi(x)\Big(1+\frac{P_2(x)}n\Big)+O\Big(\frac1{n^{3/2}}\Big) \end{equation*} uniformly in all real $x$, where $P_2$ is a certain polynomial. It follows that $I_{n1}=O(1/n)$. Also, $I_{n2}=0$, since $|S_n|\le\sqrt{\frac3n}\,n=\sqrt{3n}$. Finally, it is easy to see that $I_{n3}=o(1/n)$.

Thus, by \eqref{1},
\begin{equation*} \dee(D_n,N(0,1))=O(1/n), \end{equation*} as desired.

In fact, $P_2(x)=\frac1{20}(3-6x^2+x^4)$, and hence, slightly modifying the above reasoning, we see that \begin{equation*} \dee(D_n,N(0,1))\sim\frac cn, \end{equation*} where \begin{equation} c:=\int\vpi|P_2|= \frac{e^{-3/2-\sqrt{3/2}}}{5\sqrt\pi} \Big(e^{\sqrt{6}} \sqrt{9-3\sqrt{6}}+\sqrt{9+3\sqrt{6}}\,\Big) \\ =0.140030\ldots. \end{equation}

Here is the graph $\big\{\big(n,\frac nc\,\dee(D_n,N(0,1))\big)\colon n\in\{3,\dots,15\}\big\}$:

$\newcommand\dee{\Delta_{\mathrm{TV}}}\newcommand{\vpi}{\varphi}$Let $f_n$ denote the pdf of \begin{equation*} S_n:=\sqrt{\frac3n} \sum_{i=1}^n X_i \end{equation*} and let $\vpi$ denote the standard normal pdf. Then \begin{equation*} \dee(D_n,N(0,1))=\int|f_n-\vpi|\le I_{n1}+I_{n2}+I_{n3}, \tag{1}\label{1} \end{equation*} where \begin{equation*} I_{n1}:=\int_{|x|\le\sqrt{3n}}|f_n(x)-\vpi(x)|\,dx, \end{equation*} \begin{equation*} I_{n2}:=\int_{|x|>\sqrt{3n}}f_n(x)\,dx, \end{equation*} \begin{equation*} I_{n3}:=\int_{|x|>\sqrt{3n}}\vpi(x)\,dx. \end{equation*}

We bound $I_{n1}$ using the asymptotic expansion in the central limit theorem given (say) by Theorem 7 (with $k=5$ and $l=1$) on p. 175 in the book by Petrov. Noting also Lemma 10 on p. 173 and the expressions for $Q_{kn}(x)$ on p. 138 of the same book, as well as the fact that $EX_i=EX_i^3=0$ for all $i$, we see that \begin{equation*} f_n(x)=\vpi(x)\Big(1+\frac{P_2(x)}n\Big)+O\Big(\frac1{n^{3/2}}\Big) \end{equation*} uniformly in all real $x$, where $P_2$ is a certain polynomial. It follows that $I_{n1}=O(1/n)$. Also, $I_{n2}=0$, since $|S_n|\le\sqrt{\frac3n}\,n=\sqrt{3n}$. Finally, it is easy to see that $I_{n3}=o(1/n)$.

Thus, by \eqref{1},
\begin{equation*} \dee(D_n,N(0,1))=O(1/n), \tag{2}\label{2} \end{equation*} as desired.

In fact, $P_2(x)=\frac1{20}(3-6x^2+x^4)$, and hence, slightly modifying the above reasoning, we see that \begin{equation*} \dee(D_n,N(0,1))\sim\frac cn, \end{equation*} where \begin{equation} c:=\int\vpi|P_2|= \frac{e^{-3/2-\sqrt{3/2}}}{5\sqrt\pi} \Big(e^{\sqrt{6}} \sqrt{9-3\sqrt{6}}+\sqrt{9+3\sqrt{6}}\,\Big) \\ =0.140030\ldots. \end{equation}

Here is the graph $\big\{\big(n,\frac nc\,\dee(D_n,N(0,1))\big)\colon n\in\{3,\dots,15\}\big\}$:

Following the lines of the proof, it easy to see that \eqref{2} will hold whenever, say, the $X_i$'s are iid with $EX_i=EX_i^3=0$ and light enough distribution tails (in your case, the $X_i$'s are iid symmetric random variables with no distribution tails).

$\newcommand\dee{\Delta_{\mathrm{TV}}}\newcommand{\vpi}{\varphi}$Let $f_n$ denote the pdf of \begin{equation*} S_n:=\sqrt{\frac3n} \sum_{i=1}^n X_i \end{equation*} and let $\vpi$ denote the standard normal pdf. Then \begin{equation*} \dee(D_n,N(0,1))=\int|f_n-\vpi|\le I_{n1}+I_{n2}+I_{n3}, \tag{1}\label{1} \end{equation*} where \begin{equation*} I_{n1}:=\int_{|x|\le\sqrt{3n}}|f_n(x)-\vpi(x)|\,dx, \end{equation*} \begin{equation*} I_{n2}:=\int_{|x|>\sqrt{3n}}f_n(x)\,dx, \end{equation*} \begin{equation*} I_{n3}:=\int_{|x|>\sqrt{3n}}\vpi(x)\,dx. \end{equation*}

We bound $I_{n1}$ using the asymptotic expansion in the central limit theorem given (say) by Theorem 7 (with $k=5$ and $l=1$) on p. 175 in the book by Petrov. Noting also Lemma 10 on p. 173 and the expressions for $Q_{kn}(x)$ on p. 138 of the same book, as well as the fact that $EX_i=EX_i^3=0$ for all $i$, we see that \begin{equation*} f_n(x)=\vpi(x)\Big(1+\frac{P_2(x)}n\Big)+O\Big(\frac1{n^{3/2}}\Big) \end{equation*} uniformly in all real $x$, where $P_2$ is a certain polynomial. It follows that $I_{n1}=O(1/n)$. Also, $I_{n2}=0$, since $|S_n|\le\sqrt{\frac3n}\,n=\sqrt{3n}$. Finally, it is easy to see that $I_{n3}=O(1/n)$.

Thus, by \eqref{1},
\begin{equation*} \dee(D_n,N(0,1))=O(1/n), \end{equation*} as desired.

In fact, $P_2(x)=\frac1{20}(3-6x^2+x^4)$, and hence the above reasoning shows that \begin{equation*} \dee(D_n,N(0,1))\sim c/n, \end{equation*} where \begin{equation} c:=\int\vpi|P_2|= \frac{e^{-3/2-\sqrt{3/2}}}{5\sqrt\pi} \Big(e^{\sqrt{6}} \sqrt{9-3\sqrt{6}}+\sqrt{9+3\sqrt{6}}\,\Big) \\ =0.140030\ldots. \end{equation}

Here is the graph $\big\{\big(n,\frac nc\,\dee(D_n,N(0,1))\big)\colon n\in\{3,\dots,15\}\big\}$:

$\newcommand\dee{\Delta_{\mathrm{TV}}}\newcommand{\vpi}{\varphi}$Let $f_n$ denote the pdf of \begin{equation*} S_n:=\sqrt{\frac3n} \sum_{i=1}^n X_i \end{equation*} and let $\vpi$ denote the standard normal pdf. Then \begin{equation*} \dee(D_n,N(0,1))=\int|f_n-\vpi|\le I_{n1}+I_{n2}+I_{n3}, \tag{1}\label{1} \end{equation*} where \begin{equation*} I_{n1}:=\int_{|x|\le\sqrt{3n}}|f_n(x)-\vpi(x)|\,dx, \end{equation*} \begin{equation*} I_{n2}:=\int_{|x|>\sqrt{3n}}f_n(x)\,dx, \end{equation*} \begin{equation*} I_{n3}:=\int_{|x|>\sqrt{3n}}\vpi(x)\,dx. \end{equation*}

We bound $I_{n1}$ using the asymptotic expansion in the central limit theorem given (say) by Theorem 7 (with $k=5$ and $l=1$) on p. 175 in the book by Petrov. Noting also Lemma 10 on p. 173 and the expressions for $Q_{kn}(x)$ on p. 138 of the same book, as well as the fact that $EX_i=EX_i^3=0$ for all $i$, we see that \begin{equation*} f_n(x)=\vpi(x)\Big(1+\frac{P_2(x)}n\Big)+O\Big(\frac1{n^{3/2}}\Big) \end{equation*} uniformly in all real $x$, where $P_2$ is a certain polynomial. It follows that $I_{n1}=O(1/n)$. Also, $I_{n2}=0$, since $|S_n|\le\sqrt{\frac3n}\,n=\sqrt{3n}$. Finally, it is easy to see that $I_{n3}=o(1/n)$.

Thus, by \eqref{1},
\begin{equation*} \dee(D_n,N(0,1))=O(1/n), \end{equation*} as desired.

In fact, $P_2(x)=\frac1{20}(3-6x^2+x^4)$, and hence, slightly modifying the above reasoning, we see that \begin{equation*} \dee(D_n,N(0,1))\sim\frac cn, \end{equation*} where \begin{equation} c:=\int\vpi|P_2|= \frac{e^{-3/2-\sqrt{3/2}}}{5\sqrt\pi} \Big(e^{\sqrt{6}} \sqrt{9-3\sqrt{6}}+\sqrt{9+3\sqrt{6}}\,\Big) \\ =0.140030\ldots. \end{equation}

Here is the graph $\big\{\big(n,\frac nc\,\dee(D_n,N(0,1))\big)\colon n\in\{3,\dots,15\}\big\}$: