I encountered this problem
If $f$ is a is continuously differentiable on $[0,1]$, find $$\lim_{n \to \infty }n\left(\sum_{k=1}^n \frac{1}{n}f\left(\frac k n\right)-\int_0^1f(x)dx\right)$$
My attempt :-
for all $\varepsilon >0$ choose $N$ st $|f'(x)- f'(y)|< \frac{1}{2\varepsilon}$ when $|x-y|< 1/N $ $$L:=\lim_{n \to \infty }n\left(\sum_{k=1}^n \frac{1}{n}f\left(\frac k n\right)-\int_0^1f(x)dx\right)$$ $$-L= \lim_{n \to \infty }n\left(\sum_{k=1}^n \int_{\frac{k-1}{n}}^ {\frac{k}{n}} f(x) - f\left(\frac k n\right)dx\right)$$ $$\lim_{n \to \infty }n\left(\sum_{k=1}^n \int_{\frac{k-1}{n}}^ {\frac{k}{n}} f(x) -f\left(\frac {k-1} n\right) + f\left(\frac {k-1} n\right) - f\left(\frac k n\right)dx\right)$$ $${\lim_{n \to \infty }\left(\sum_{k=1}^n \int_{\frac{k-1}{n}}^ {\frac{k}{n}} \frac {f(x) -f\left(\frac {k-1} n\right)}{1/n} + \frac{f\left(\frac {k-1} n\right) - f\left(\frac k n\right)}{1/n}dx\right)}$$ $$\lim_{n \to \infty }\left(\sum_{k=1}^n \int_{\frac{k-1}{n}}^ {\frac{k}{n}} f'\left(\frac{k-1}{n}\right)\left(\frac{1}{2}\right)dx \right)-\varepsilon < L< \lim_{n \to \infty }\left(\sum_{k=1}^n \int_{\frac{k-1}{n}}^ {\frac{k}{n}} f'\left(\frac{k-1}{n}\right)\left(\frac{1}{2}\right)dx \right)+\varepsilon $$ $$L = \frac1 2 \int_0 ^1 f'(x)dx = \frac{f(1)-f(0)}{2}$$
Now my question is: can we generalise this ?
For $f:[0,1] \to R$, $f^{(m)}$ is continuous, lets define $\displaystyle l_0:= \int_0^1 f(x)dx$, $\displaystyle x_{n,0}:=\sum_{k=1}^n \frac{1}{n}f\left(\frac k n\right)$, $\displaystyle x_{n , m+1}= n (x_{n,m}- l_{m} )$ and $\displaystyle l_{m}= \lim\limits_{n \to \infty} x_{n,m}$.
Assuming of course that $l_m \in \mathbb{R}$
How to determine $l_m $ ?
