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Show that if $f$ is continuously differentiable on $[0,1]$, then $$\lim\limits_{n\rightarrow\infty}n\left(\frac{1}{n}\sum_{i=1}^{n}f\left(\frac{i}{n}\right)-\int_{0}^{1}f(x)dx\right)=\frac{f(1)-f(0)}{2}$$


Observe that \begin{align*} \lim\limits_{n\rightarrow\infty}n\left(\frac{1}{n}\sum_{i=1}^{n}f\left(\frac{i}{n}\right)-\int_{0}^{1}f(x)dx\right)&=\lim\limits_{n\rightarrow\infty}n\left(\frac{1}{n}\sum_{i=1}^{n}f\left(\frac{i}{n}\right)-\sum_{i=1}^{n}\int_{(i-1)/n}^{i/n}f(x)dx\right)\\ &=\lim\limits_{n\rightarrow\infty}n\left(\sum_{i=1}^{n}\int_{(i-1)/n}^{i/n}\left[f\left(\frac{i}{n}\right)-f(x)\right]dx\right)\\ &=\lim\limits_{n\rightarrow\infty}n\left(\sum_{i=1}^{n}\int_{(i-1)/n}^{i/n}f'(c_i)\left[\left(\frac{i}{n}\right)-x\right]dx\right) \end{align*} where the last equality follows from the Mean Value Theorem.

Let $m_i=\inf\{f'(x):x\in[(i-1)/n,i/n]\}$ and $M_i=\sup\{f'(x):x\in[(i-1)/n,i/n]\}$, then we have the follow inequality: $$m_i\int_{(i-1)/n}^{i/n}\left[\left(\frac{i}{n}\right)-x\right]dx\leq\int_{(i-1)/n}^{i/n}f'(c_i)\left[\left(\frac{i}{n}\right)-x\right]dx\leq M_i\int_{(i-1)/n}^{i/n}\left[\left(\frac{i}{n}\right)-x\right]dx$$

Consequently $$\frac{1}{2n}\sum_{i=1}^{n}m_i\leq n\sum_{i=1}^{n}\int_{(i-1)/n}^{i/n}f'(c_i)\left[\left(\frac{i}{n}\right)-x\right]dx\leq\frac{1}{2n}\sum_{i=1}^{n} M_i$$ where $\int_{(i-1)/n}^{i/n}\left[\left(\frac{i}{n}\right)-x\right]dx=\frac{1}{2n^2}$.


I stuck at this step. And it seems not right because when I take the limit both sides, I have $0$. Can someone give me a hint or suggestion. Thanks in advanced.

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  • $\begingroup$ You can't just write $f'(c_i)$, since $c_i$ depends on $x$ - it is a different value for each $x$. $\endgroup$ – Thomas Andrews Dec 23 '15 at 0:39
  • $\begingroup$ @ThomasAndrews that equality wrong? $\endgroup$ – Simple Dec 23 '15 at 0:52
  • $\begingroup$ Yes, your application of the mean value theorem is wrong - otherwise, every function is linear if there is a $c\in (a,b)$ such that $f(x)-f(a)=f'(c)(x-a)$ for all $x\in(a,b)$. $\endgroup$ – Thomas Andrews Dec 23 '15 at 0:55
  • $\begingroup$ @ThomasAndrews Can you give me a suggestion to modify it? $\endgroup$ – Simple Dec 23 '15 at 1:07
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    $\begingroup$ Possible duplicate of Limit with Integral and Sigma $\endgroup$ – Paramanand Singh Jun 13 '16 at 5:49
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Your bounds

$$ \frac{1}{2n} \sum_{i=1}^{n} m_i \quad \text{and} \quad \frac{1}{2n} \sum_{i=1}^{n} M_i $$

converge to the same quantity, namely $\frac{1}{2}(f(1) - f(0))$. This is essentially because they are Riemann sums for $\frac{1}{2}f'(x)$.


Proof using Taylor Theorem. Let $x_i = i/n$ for brevity and consider $F(x) = \int_{0}^{x} f(t) \, dt$. Then we may write

$$ n \left( \frac{1}{n} \sum_{i=1}^{n} f(x_i) - \int_{0}^{1} f(x) \, dx \right) = n \sum_{i=1}^{n} (F(x_{i-1}) - F(x_i) + \tfrac{1}{n}F'(x_i)). $$

By Taylor Theorem, we can pick $c_i \in [x_{i-1}, x_i]$ such that

$$ F(x_{i-1}) = F(x_i - \tfrac{1}{n}) = F(x_i) - \tfrac{1}{n}F'(x_i) + \tfrac{1}{2n^2}F''(c_i). $$

Plugging this back, we have

$$ n \left( \frac{1}{n} \sum_{i=1}^{n} f(x_i) - \int_{0}^{1} f(x) \, dx \right) = \frac{1}{2n} \sum_{i=1}^{n} f'(c_i). $$

Taking $n \to \infty$, this converges to $\frac{1}{2}(f(1) - f(0))$ as desired.

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Hint: Use Taylor's theorem and write the following: $$f(x) = f(\frac{i}{n}) +f'(\frac{i}{n})(x-\frac{i}{n})+h_i(x)(x-\frac{i}{n}) $$, where $h_i(x)$ is a continuous function that has a limit 0 at $x = \frac{i}{n}$. The rest should be a matter of computation.

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  • $\begingroup$ I don't know the Taylow's theorem yet $\endgroup$ – Simple Dec 23 '15 at 1:08
  • $\begingroup$ Then, I think the answer by @sangchul lee serves the purpose. $\endgroup$ – dezdichado Dec 23 '15 at 3:12
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$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Leftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\, #2 \,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ Indeed, it involves the application of Bernoulli Numbers $\braces{B_{i}}$: \begin{align} \sum_{i = 1}^{n}\mathrm{f}\pars{{i \over n}} & = \int_{0}^{n}\mathrm{f}\pars{{x \over n}}\,\dd x + \sum_{i = 1}^{m}{B_{i} \over i!\,n^{i - 1}}\bracks{% \mathrm{f}^{\pars{i - 1}}\pars{1} - \mathrm{f}^{\pars{i - 1}}\pars{0}} + R_{+}\pars{\mathrm{f},m} \\[3mm] & = n\int_{0}^{1}\mathrm{f}\pars{x}\,\dd x\ +\ \overbrace{B_{1}}^{\ds{\half}}\ \bracks{\mathrm{f}\pars{1} - \mathrm{f}\pars{0}} + {1 \over 2n}\ \overbrace{B_{2}}^{\ds{{1 \over 6}}}\bracks{\mathrm{f}'\pars{1} - \mathrm{f}'\pars{0}} + R_{+}\pars{\mathrm{f},2} \end{align}

which yields \begin{align} &n\bracks{{1 \over n}\sum_{i = 1}^{n}\mathrm{f}\pars{{i \over n}} - \int_{0}^{n}\mathrm{f}\pars{{x \over n}}\,\dd x} = {\mathrm{f}\pars{1} - \mathrm{f}\pars{0} \over 2} + {\mathrm{f}'\pars{1} - \mathrm{f}'\pars{0} \over 12n} + R_{+}\pars{\mathrm{f},2} \end{align}


Then, \begin{align} \color{#f00}{\lim_{n \to \infty}n\bracks{% {1 \over n}\sum_{i = 1}^{n}\mathrm{f}\pars{{i \over n}} - \int_{0}^{1}\mathrm{f}\pars{x}\,\dd x}} & = \color{#f00}{{\mathrm{f}\pars{1} - \mathrm{f}\pars{0} \over 2}} \end{align}

whenever the remaining term $R_{+}\pars{\mathrm{f},2}$ vanishes out ( see the above cited link ) in the $n \to \infty$ limit.

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