Consider $f:[0,1]\rightarrow \mathbb{R}$ and define \begin{equation} \sigma_n(f):=\frac{1}{n}\sum_{k=1}^{n}f \left(\frac{k}{n}\right) \end{equation} for $n=1,2,\dots$
I'm trying to prove that if $f\in C^1([0,1])$, then there exists a constant (depending on the function $f$) $a(f)$ such that \begin{equation} \sigma_n(f)=a+\frac{\sigma_n(f')}{2n}+o\left(\frac{1}{n}\right). \end{equation}
I have tried to use Taylor' series about $\frac{k-1}{n}$ for $k=1,\dots,n$ but I can't figure it out...it seems to be not working...
It appears from the comment thread that we have an idea how to prove the result, but there's confusion over why any correct proof must use the fact that $f'$ is continuous. First we give a counterexample: A differentiable function $f$, with $f'$ continuous except at the origin, such that the conclusion is false. Then for the sake of tidiness we give a proof of the result assuming $f'$ is continuous.
At first I thought about showing that the conclusion fails for the traditional $t^2\sin(1/t)$ thing, but estimating the required $\sigma_n(f)$ and $\sigma_n(f')$ for that function made my head hurt. Decided to actually write down an example when I realized the calculations could be much simplified by building $f$ out of non-decreasing functions.
Let $\phi:\Bbb R\to\Bbb R$ be smooth and non-decreasing, with $\phi(t)=0$ for $t<-1$, $\phi(t)=1$ for $t>1$, and $\phi'(0)=1$. Let $$f_n(t)=\phi\left(100^n(t-2^{-n})\right),$$and define $$f=\sum_{n=1}^\infty 4^{-n}f_n.$$
First, $f$ is certainly continuous, so $$\sigma_n(f)\to\int_0^1f(t)\,dt.$$
Note that $f_n'=0$ on $\Bbb R\setminus I_n$, where $I_n=(2^{-n}-100^{-n},2^{-n}+100^{-n})$. Since the $I_n$ are disjoint it is clear that $f$ is differentiable except perhaps at the origin where the $I_n$ pile up, and that in fact $f'$ is continuous away from the origin. Now $$0\le f_n\le\chi_{[2^{-n}-100^{-n},\infty)};$$this shows that $$0\le f(t)\le ct^2$$near the origin, so $f$ is differentiable at the origin. So $f$ is differentiable.
Since $f_n'\ge0$ it follows that $$\sigma_{2^n}(f')\ge 2^{-n}f'(2^{-n}) \ge 8^{-n}f_n'(2^{-n})=(100/8)^n.$$So $$2^{-n}\sigma_{2^n}(f')\to\infty,$$showing that $\lim\frac1{2n}\sigma_n(f')$ does not exist.
Proof, using continuity of $f'$:
Let $\epsilon>0$. Since $f'$ is uniformly continuous there exists $N$ such that $$|f'(t)-f'(s)|<\epsilon\quad(|t-s|<1/N).$$
Assume that $n>N$. Let $a=\int_0^1 f(t)\,dt$. Now
$$\sigma_n(f)-a=\sum_{k=1}^n\int_{(k-1)/n}^{k/n}(f(k/n)-f(t))\,dt.$$If $t\in[(k-1)/n,k/n]$ then $$\begin{align}f(k/n)-f(t)&=\int_t^{k/n}f'(s)\,ds \\&=(k/n-t) f'(k/n)+\int_t^{k/n}(f'(s)-f'(k/n))\,ds\\&=(k/n-t)f'(k/n)+E_{n,k}(t),\end{align}$$where $$|E_{n,k}(t)|<\frac\epsilon n.$$So $$\int_{(k-1)/n}^{k/n} (f(k/n)-f(t)) =\frac1{2n^2}f'(k/n)+\int_{(k-1)/n}^{k/n}E_{n,k}(t)\,dt.$$Inserting this above shows that $$\sigma_n(f)-a=\frac1{2n}\sigma_n(f')+\sum_{k=1}^n\int_{(k-1)/n}^{k/n}E_{n,k}(t)\,dt$$and$$\left|\sum_{k=1}^n\int_{(k-1)/n}^{k/n}E_{n,k}(t)\,dt\right|<\frac\epsilon n.$$