Limit of a Riemann Sum and Integral I've been trying to solve this problem, but I haven't been able to calculate the exact limit, I've just been able to find some boundaries. I hope you guys can help me with it.
Let $f:[0,1] \to \mathbb{R}$ a differentiable function with a continuous derivative, calculate:
$$\lim_{n\to \infty}\left(\sum_{k=1}^nf\left(\frac{k}{n}\right)-n\int_0^1f(x)dx\right) $$
I tried using Mean Value Theorem for derivatives and integrals and I got that 
$$\lim_{n\to \infty}\left(\sum_{k=1}^nf\left(\frac{k}{n}\right)-n\int_0^1f(x)dx\right)=\lim_{n\to \infty}\left(\sum_{k=1}^nf'\left(x_k**\right)(\frac{k}{n}-x_k*)\right) $$
Where $x_k*\in [\frac{k-1}{n},\frac{k}{n}]$ and $x_k**\in [x_k*,\frac{k}{n}]$, which looks like a Riemann Sum but I'm not sure if it's a Riemann Sum of $f'$ from $0$ to $1$, if this was true I believe the limit is $f(1)-f(0)$ but I'm not really sure about this.
Edit: fixed some typos with the $\frac{k}{n}$.
Edit 2: $k$ starts from $1$ not $0$.
 A: Let
$$
x_n=\sum_{k=0}^nf\left(\frac{k}{n}\right)-n\int_0^1f(x)dx
$$
We will use the following result:
Lemma If $g:[0,1]\to\mathbb{R}$ is a continuously differentiable function. Then
$$
 \frac{g(0)+g(1)}{2}-\int_0^1g(x)dx= \int_{0}^1\left(x-\frac{1}{2}\right)g'(x)dx.
$$
Indeed, this is just integration by parts:
$$\eqalign{
\int_{0}^1\left(x-\frac{1}{2}\right)g'(x)dx
&=\left.\left(x-\frac{1}{2}\right)g(x)\right]_{x=0}^{x=1}
-\int_0^1g(x)dx\cr
&=\frac{g(1)+g(0)}{2}-\int_0^1g(x)dx
}$$
Now applying this to the functions
$x\mapsto f\left(\frac{k+x}{n}\right)$ for $k=0,1,\ldots,n-1$ and adding the resulting inequalities we obtain
$$
x_n-\frac{f(0) +f(1)}{2} = \int_0^1\left(x-\frac{1}{2}\right)H_n(x)dx\tag{1}
$$
where,
$$
H_n(x)=\frac{1}{n}\sum_{k=0}^{n-1}f'\left(\frac{k+x}{n}\right)
$$
Clearly for every $x$, $H_n(x)$ is a Riemann sum of the function continuous  $f'$, hence
$$
\forall\,x\in[0,1],\quad\lim_{n\to\infty}H_n(x)=\int_0^1f'(t)dt
$$
 Moreover, $| H_n(x)|\leq\sup_{[0,1]}|f'|$. So,
 taking the limit in $(1)$ and 
 applying the Dominated Convergence Theorem, we obtain
$$
\lim_{n\to\infty}\left(x_n-\frac{f(0) +f(1)}{2}\right)=
\left(\int_0^1f'(t)dt\right)\int_0^1\left(x-\frac{1}{2}\right)dx=0.
$$
This proves that
$$
\lim_{n\to\infty}x_n=\frac{f(0) +f(1)}{2}
$$
And consequently
$$
\lim_{n\to\infty}\left(\sum_{k=\color{red}{1}}^nf\left(\frac{k}{n}\right)-n\int_0^1f(x)dx\right)=\frac{f(1)-f(0)}{2}
$$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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 \newcommand{\dd}{{\rm d}}
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 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
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$\ds{\lim_{n\ \to\ \infty}\bracks{\sum_{k = 1}^{n}\fermi\pars{k \over n}
    -n\int_{0}^{1}\fermi\pars{x}\,\dd x}:\ {\large ?}}$.

$\ds{\tt\mbox{This is an application of}}$
  Abel-Plana formula:

\begin{align}&\color{#c00000}{\sum_{k = 1}^{n}\fermi\pars{k \over n}}
=\sum_{k = 0}^{n - 1}\fermi\pars{k + 1 \over n}
=\sum_{k = 0}^{\infty}\bracks{\fermi\pars{k + 1 \over n}
-\fermi\pars{k + n + 1 \over n}}
\\[5mm]&=\int_{0}^{\infty}
\bracks{\fermi\pars{x + 1 \over n} - \fermi\pars{x + n + 1 \over n}}\,\dd x
+\half\bracks{\fermi\pars{1 \over n} - \fermi\pars{n + 1 \over n}}
\\[3mm]&+\color{#00f}{\ic\int_{0}^{\infty}
\bracks{\fermi\pars{\ic x + 1 \over n} - \fermi\pars{\ic x + n + 1 \over n}
-\fermi\pars{-\ic x + 1 \over n} + \fermi\pars{-\ic x + n + 1 \over n}}\times}
\\[3mm]&\color{#00f}{\dd x \over \expo{2\pi x} - 1}
\\[5mm]&=n\int_{1/n}^{\infty}\fermi\pars{x}\,\dd x
-\int_{1 + 1/n}^{\infty}\fermi\pars{x}\,\dd x
+\half\bracks{\fermi\pars{1 \over n} - \fermi\pars{n + 1 \over n}}
+ \color{#00f}{"\mbox{the blue term}"}
\\[3mm]&=n\int_{1/n}^{1 + 1/n}\fermi\pars{x}\,\dd x
+\half\bracks{\fermi\pars{1 \over n} - \fermi\pars{n + 1 \over n}}
+ \color{#00f}{"\mbox{the blue term}"}
\\[3mm]&=n\int_{0}^{1}\fermi\pars{x}\,\dd x
-n\int_{0}^{1/n}\fermi\pars{x}\,\dd x + n\int_{1}^{1 + 1/n}\fermi\pars{x}\,\dd x
+\half\bracks{\fermi\pars{1 \over n} - \fermi\pars{n + 1 \over n}}
\\[3mm]&\mbox{}+ \color{#00f}{"\mbox{the blue term}"}
\end{align}

Since $\ds{\lim_{n\ \to\ \infty}\color{#00f}{\pars{"\mbox{the blue term}"}} = 0}$
  and $\ds{\lim_{n\ \to\ \infty}n\int_{0}^{1/n}\fermi\pars{x}\,\dd x = \fermi\pars{0}}$
  and $\ds{\lim_{n\ \to\ \infty}n\int_{1}^{1 + 1/n}\fermi\pars{x}\,\dd x = \fermi\pars{1}}$:

\begin{align}
&\color{#66f}{\large\lim_{n\ \to\ \infty}\bracks{%
\sum_{k = 1}^{n}\fermi\pars{k \over n} - n\int_{0}^{1}\fermi\pars{x}\,\dd x}}
\\[3mm]&=-\fermi\pars{0} + \fermi\pars{1}
+\half\bracks{\fermi\pars{0} - \fermi\pars{1}}
=\color{#66f}{\large{\fermi\pars{1} - \fermi\pars{0} \over 2}}
\end{align}
A: This was brought up recently in chat and I wrote this up.  This question seemed like a good home.

As hinted in the following diagram

$$
\int_{\frac{k-1}n}^{\frac{k}n}(f(k/n)-f(t))\,\mathrm{d}t\sim\frac{f'(k/n)}{2n^2}\tag1
$$
In fact, integration by parts and the mean value theorem yield for some $\xi\in\left(\frac{k-1}n,\frac{k}n\right)$:
$$
\begin{align}
\int_{\frac{k-1}n}^{\frac{k}n}(f(k/n)-f(t))\,\mathrm{d}t
&=\int_{\frac{k-1}n}^{\frac{k}n}\left(t-\frac{k-1}n\right)f'(t)\,\mathrm{d}t\tag{2a}\\
&=\frac{f'(k/n)}{2n^2}-\int_{\frac{k-1}n}^{\frac{k}n}\frac12\left(t-\frac{k-1}n\right)^2f''(t)\,\mathrm{d}t\tag{2b}\\
&=\frac{f'(k/n)}{2n^2}-\frac1{6n^3}f''(\xi)\tag{2c}
\end{align}
$$
Explanation:
$\text{(2a)}$: Integration by Parts
$\text{(2b)}$: Integration by Parts
$\text{(2c)}$: Mean Value Theorem
Therefore,
$$
\begin{align}
\lim_{n\to\infty}\left(\sum_{k=1}^nf(k/n)-n\int_0^1f(t)\,\mathrm{d}t\right)
&=\lim_{n\to\infty}\left(n\sum_{k=1}^n\int_{\frac{k-1}n}^{\frac{k}n}(f(k/n)-f(t))\,\mathrm{d}t\right)\tag{3a}\\
&=\lim_{n\to\infty}\sum_{k=1}^n\frac12f'(k/n)\,\frac1n\tag{3b}\\
&=\frac12(f(1)-f(0))\tag{3c}
\end{align}
$$
Explanation:
$\text{(3a)}$: break up the integral and move the sum inside the integral
$\text{(3b)}$: apply $(2)$
$\text{(3c)}$: apply Riemann Summation
A: From the error analysis paragraph in the Wikipedia page for the Trapezoidal rule we have that the limit is: $$\frac{f(0)+f(1)}{2}.$$
