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We have, $$f(b)-f(a)=\lim_{n\rightarrow \infty}\sum_{k=0}^{n-1} hf'(a+kh)\:\:\:\:\:..(1)$$

where $h=\frac{b-a}{n}$

Now, $$f'(a)=f'(a)$$

$$f'(a+h)=f'(a)+hf^2(a)$$ $f^2(a)$ meaning the second derivative at $a$

$$f'(a+2h)=f'(a+h)+hf^2(a+h)=f'(a)+hf^2(a)+h(f^2(a)+hf^3(a))$$ $$=f'(a)+2hf^2(a)+h^2f^3(a)$$

Similarly, I found $f'(a+3h)$ to be $$f'(a)+3hf^2(a)+3h^2f^3(a)+h^3f^4(a)$$

Observing the pattern,

$$f'(a+kh)=\sum_{r=0}^k \binom{k}{r} h^rf^{r+1}(a)$$

So, the expansion of $\sum_{k=0}^{n-1} f'(a+kh)$ will contain terms of the form $c_rh^r f^{r+1}(a)$ with $r$ ranging from $0$ to $n-1$ and the coefficient $c_r$ of $h^r f^{r+1}(a)$ being $\sum_{k=0}^{n-1}\binom{k}{r}$

Therefore, $$f(b)-f(a)=\lim_{n\rightarrow \infty}h\sum_{r=0}^{n-1} \left(h^{r}f^{r+1}(a)\sum_{k=0}^{n-1} \binom{k}{r}\right)$$

$$=\lim_{n\rightarrow \infty}\sum_{r=0}^{n-1} \left(\frac{b-a}{n}\right)^{r+1}f^{r+1}(a)\binom{n}{r+1}$$

Compare this with the Taylor series result where

$$f(b)-f(a)=\lim_{n\rightarrow \infty} \sum_{r=0}^n (b-a)^{r+1} f^{r+1}(a)\frac{1}{(r+1)!}$$

Is the result similar or equivalent to the Taylor series? Also, is there some formula in terms of $n$ and $r$ for $\sum_{k=0}^{n-1} \binom{k}{r}$?

EDIT:

I think

$$\lim_{n\rightarrow \infty}\sum_{r=0}^{n-1} \left(\frac{b-a}{n}\right)^{r+1}f^{r+1}(a)\binom{n}{r+1}$$

$$=\lim_{n\rightarrow \infty}\sum_{r=0}^{n-1} \left(\frac{b-a}{n}\right)^{r+1}f^{r+1}(a)\frac{n!}{(n-(r+1))!(r+1)!}$$ might be equivalent to

$$\lim_{n\rightarrow \infty} \sum_{r=0}^n (b-a)^{r+1} f^{r+1}(a)\frac{1}{(r+1)!}$$

because of the limit. Is this true?

EDIT: Would this post be more rigorous if I removed all the limits and used $\omega$ in place of $n$? $\omega$ is the infinity in hyperreal numbers. $\frac{1}{\omega}$ is the infinitesimal. Arithmetic with infinity and infinitesimals is allowed in hyper-reals. That way it'd be non-standard but pretty rigorous, I think? I was using limits just to make it look as much mainstream as possible.