# Riemann Integral : $\lim_{n\rightarrow \infty}\frac{1}{n}\sum_{k=1}^{n}f'(\frac{k}{3n})$

Let f $\in$ $C^1[-1,1]$. Evaluate

$$\lim_{n\rightarrow \infty}\frac{1}{n}\sum_{k=1}^{n}f'(\frac{k}{3n})$$

My approach

I have tried to convert the summation into a Riemann Integral as follows:

\begin{align*} \implies 3\cdot \lim_{n\rightarrow \infty} \sum_{k=1}^{n}f'\left(\frac{k}{3n}\right)\cdot \left( \frac{k}{3n}-\frac{k-1}{3n}\right) \end{align*}

By partitioning the interval $[0,1/3]$ into $(n-1)$ intervals. But how do I show that this is the Riemann Sum that corresponds to the integral:

$$\int_0^{1/3}f'(x)dx$$

• Let $g(x) = f'(x/3)$, then the limit is $\int_0^1 g(x)dx = \int_0^1 f'(x/3)dx$, now do a change of variable. – user384138 Jan 20 '17 at 14:50
• @OpenBall How do I show that the limit is $\int_0^1 g(x) dx$ after I substituted $g(x)=f'(x/3)$ in the last step I showed. – mm-crj Jan 20 '17 at 15:06

You already have a Riemann sum (or almost). Let $g\colon[-3,3]\to\mathbb{R}$ be the $\mathcal{C}^1$ function defined by $g(x) = f(\frac{x}{3})$. Then $g'(x) = \frac{1}{3}f'(\frac{x}{3})$ for all $x\in[-1,1]$, and for $n\geq 1$ \begin{align} \frac{1}{n}\sum_{k=1}^n f'\left(\frac{k}{3n}\right) &= 3\cdot \frac{1}{n}\sum_{k=1}^n \frac{1}{3}f'\left(\frac{k}{3n}\right) = 3\cdot \frac{1}{n}\sum_{k=1}^n g'\left(\frac{k}{n}\right)\\&\xrightarrow[n\to\infty]{} 3\int_0^1 g'(x)dx = 3\left(g(1)-g(0)\right) = 3\left(f\left(\frac{1}{3}\right)-f(0)\right) \end{align}
• $f$ being the identity function: then, \begin{align} \frac{1}{n}\sum_{k=1}^n f'\left(\frac{k}{3n}\right) &= \frac{1}{n}\sum_{k=1}^n 1 = 1 \\&\xrightarrow[n\to\infty]{} 1= \boxed{3\left(f\left(\frac{1}{3}\right)-f(0)\right)} \end{align} indeed.
• $f$ being the the square function, $f(x)=x^2$. Then \begin{align} \frac{1}{n}\sum_{k=1}^n f'\left(\frac{k}{3n}\right) &= \frac{2}{3}\cdot \frac{1}{n}\sum_{k=1}^n \frac{k}{n} \\&\xrightarrow[n\to\infty]{} \frac{2}{3}\cdot\frac{1}{2} = \frac{1}{3} = 3\left(f\left(\frac{1}{3}\right)-f(0)\right) \end{align} again.