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](http://en.wikipedia.org/wiki/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](http://en.wikipedia.org/wiki/Edgeworth_series) 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.



estimate the error term in CLT
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 $(2)$. 


