TITLE: estimate the error term in CLT
QUESTION [9 upvotes]: Let $X_m = \frac{1}{\sqrt{m}}\sum_{k=1}^m Z_k$ where $Z_k$ are iid equally likely on $\{\pm 1\}$. Then $X_m$ convergens to $X \sim \mathcal{N}(0,1)$ in distribution by CLT.
Let $f$ be a smooth bounded function on $\mathbb{R}$. Then $\mathbb{E}[f(X_m)] \to \mathbb{E}[f(X)]$. I wonder if there is any general method to give sharp asymptotic estimate of the error term $\mathbb{E}[f(X_m)] - \mathbb{E}[f(X)]$, which I expect to be $\Theta(1/m)$. The scaling constant should depend on $f$ (as well as the distribution of $Z_k$ if they are not binary). 
For law of large number, this type of estimate can be done via the Delta method (e.g., to estimate $\mathbb{E}[f(\bar{Z})] - f(0)$). There must be a counterpart for CLT... I haven't found the Edgeworth expansion useful because it seems to work with distribution with densities.
Edited: To be clear, I am only interested in some specific nice function (e.g., $f(x) = x^2 e^{-x^2/4}$) and finding a sharp expansion for the error term of the form, say, $c/m + o(1/m)$, where $c$ will depend n $f$. As pointed by Mark, the worst-case rate of all bounded smooth function $f$ is $1/\sqrt{m}$, which agrees with the upper bound given by Stein's method.

REPLY [2 votes]: Your conjecture is correct. 
Suppose that (say) $f$ has a bounded $5$th derivative. Then
\begin{equation}
 Ef(X_m) - Ef(X)=-\frac{Ef''''(X)+o(1)}{12m}. \tag{1}
\end{equation}
Indeed, let 
\begin{equation}
 Z_{mj}:=\frac{Z_j}{\sqrt m},\quad Y_{mj}:=\frac{Y_j}{\sqrt m},\quad T_{mk}:=\sum_{j=1}^{k-1}Z_{mj}+\sum_{j=k+1}^m Y_{mj}, 
\end{equation}
where $Z_1,\dots,Z_m,Y_1,\dots,Y_m$ are independent random variables and $Y_j\sim N(0,1)$. 
Note that $EZ_{mj}^p=EY_{mj}^p$ for $p=0,\dots,3$, $EZ_{mj}^4=\tfrac1{m^2}$, $EY_{mj}^4=\tfrac3{m^2}$. 
Also, $Z_{mk}$ is independent of $T_{mk}$ for each $k$, and so is $Y_{mk}$. 
Moreover, $Ef''''(T_{mk})\to Ef''''(X)$ uniformly in $k=1,\dots,m$ (as $m\to\infty$); this should be as easy to show as the convergence $Ef(X_m) \to Ef(X)$, mentioned in the question. 
Now one has
\begin{equation}
 Ef(X_m) - Ef(X)=\sum_{k=1}^m D_k, \tag{2}
\end{equation}
where
\begin{equation}
 D_k:=E[f(T_{mk}+Z_{mk})-f(T_{mk}+Y_{mk})]
\end{equation}
\begin{equation}
 =\sum_{p=0}^4\frac1{p!}\,Ef^{(p)}(T_{mk})E(Z_{mk}^p-Y_{mk}^p)+O(E(|Z_{mk}|^5+|Y_{mk}|^5))
\end{equation}
\begin{equation}
 =\tfrac1{4!}\,(Ef''''(X)+o(1))(\tfrac1{m^2}-\tfrac3{m^2})+O(m^{-5/2}) 
\end{equation}
by Taylor's expansion, 
with $o(1)$ uniform in $k=1,\dots,m$. 
Now $(1)$ follows immediately from $(2)$.