Suppose $f$ is an infinite times differentiable function whose Taylor series converges. I'm trying to prove $$\lim_{w \rightarrow \infty} \sum_{r=0}^{w - 1} \frac{f^{n+1+r}(a)}{(n+1+r)!}(x-a)^{n+1+r}=\frac{1}{n!}\int_a^x (x-t)^{n}f^{n+1}(t)dt$$
(where $f^{n+1}$ is nth derivative of $f$)
by this method:
$$RHS=\frac{1}{n!}\int_a^x (x-t)^{n}f^{n+1}(t)dt=\lim_{w \rightarrow \infty} \frac{h}{n!} \sum_{k=0}^{w-1} ((x-a)-kh))^nf^{n+1}(a+kh)$$
(where $h=\frac{x-a}{w}$)
$$=\lim_{w \rightarrow \infty} \frac{x-a}{wn!} \sum_{k=0}^{w-1} \left(\left(\sum_{r=0}^{n} \binom{n}{r} (-1)^r (x-a)^{n-r} \left(k\frac{x-a}{w}\right)^r\right)f^{n+1}(a+kh)\right)$$
$$=\lim_{w\rightarrow \infty} \frac{(x-a)^{n+1}}{n!} \sum_{k=0}^{w-1} \left(\left(\sum_{r=0}^{n} \binom{n}{r} (-1)^r \frac{k^r}{w^{r+1}}\right)f^{n+1}(a+kh)\right)$$
So, this is an expansion of $\frac{1}{n!}\int_a^x (x-t)^{n}f^{n+1}(t)dt$ with terms of the form $c_kf^{n+1}(a+kh)$ with coefficients $c_k$ being $\frac{(x-a)^{n+1}}{n!}\sum_{r=0}^{n} \binom{n}{r} (-1)^r \frac{k^r}{w^{r+1}}$
Now,
$$LHS=\lim_{w \rightarrow \infty} \sum_{r=0}^{w - 1} \frac{f^{n+1+r}(a)}{(n+1+r)!}(x-a)^{n+1+r}$$
Using $$f^{n+1+r}(a)=\lim_{h\rightarrow 0} \frac{\sum_{k=0}^{r} \binom{r}{k} (-1)^{r-k} f^{n+1} (a+kh)}{h^r}$$
$$=\lim_{w\rightarrow \infty} \frac{w^r \sum_{k=0}^{r}\binom{r}{k} (-1)^{r-k} f^{n+1} (a+kh)}{(x-a)^r}$$
$$LHS=\lim_{w \rightarrow \infty}\left( \sum_{r=0}^{w - 1} \frac{(x-a)^{n+1+r}}{{(n+1+r)!}} \left(\frac{w^r \sum_{k=0}^{r}\binom{r}{k} (-1)^{r-k} f^{n+1} (a+kh)}{(x-a)^r}\right)\right)$$
$$=(x-a)^{n+1}\lim_{w \rightarrow \infty}\left( \sum_{r=0}^{w - 1} \frac{1}{{(n+1+r)!}} \left(w^r \sum_{k=0}^{r}\binom{r}{k} (-1)^{r-k} f^{n+1} (a+kh)\right)\right)$$
This is also an expansion with terms of the form $c_k f^{n+1}(a+kh)$. For a particular value of $k$, $r$ will vary from $k$ to $w-1$ and also $k$ will vary from $0$ to $w-1$ in the entire expansion. So, $c_k=(x-a)^{n+1} \sum_{r=k}^{w-1} \binom{r}{k} \frac{ (-1)^{r-k} w^r }{(n+1+r)!}$
For LHS to be equal to RHS, the coefficients of $f^{n+1}(a+kh)$ terms must be equal on both sides. So,
$$\lim_{w\rightarrow \infty}\frac{(x-a)^{n+1}}{n!}\sum_{r=0}^{n} \binom{n}{r} (-1)^r \frac{k^r}{w^{r+1}}=(x-a)^{n+1} \lim_{w\rightarrow \infty} \sum_{r=k}^{w-1} \binom{r}{k} \frac{ (-1)^{r-k} w^r }{(n+1+r)!}$$
which implies
$$\lim_{w\rightarrow \infty}\frac{1}{n!}\sum_{r=0}^{n} \binom{n}{r} (-1)^r \frac{k^r}{w^{r+1}}=\lim_{w\rightarrow \infty} \sum_{r=k}^{w-1} \binom{r}{k} \frac{ (-1)^{r-k} w^r }{(n+1+r)!}$$
But this equation doesn't seem correct to me because LHS seems to be $0$ while RHS seems to diverge to infinity. Is this equation really correct or are there wrong steps in the proof?