Taylor series proof

$\text{Let}~ x,t,a \in \mathbb{R} ~,f(x)~\text{be a real valued function s.t. infinite times differentiable}$

and let

$\large{D^nf(x)= \frac{d^nf}{dx^n}~,~j_{n}(t)= \frac{(-1)^{n+1}(x-t)^n}{n{!}}}$

by solving this simple integral, we get 

$$\int^{x}_{a} {(x-t)~D^2f(t)}~dt = f(x)-f(a)-f^{\prime}(a)(x-a) $$

and using integration-by-parts method, we are able to observe that

$$\int^{x}_{a} {(x-t)~D^2f(t)}~dt$$

$$=\left[  j_{2}(t)D^2f(t)\right]^{x}_{{a}}-\int^{x}_{a}{j_{2}(t)D^3f(t)}dt$$

$$=\left[  j_{2}(t)D^2f(t)\right]^{x}_{{a}}-\left[  j_{3}(t)D^3f(t)\right]^{x}_{a}+\int^{x}_{a}{j_{3}(t)D^4f(t)}dt  $$

$$=\left[  j_{2}(t)D^2f(t)\right]^{x}_{{a}}-\left[  j_{3}(t)D^3f(t)\right]^{x}_{a}+\left[  j_{4}(t)D^4f(t)\right]^{x}_{a}-\int^{x}_{a}{j_{4}(t)D^5f(t)}dt ~\cdots$$

since $j_{n}(x)=0.j_{n}(a)=(x-a)^n(-1)^{n+1}$ the equation above becomes 

$$= \frac{{(x-a)^2 D^2f(a)}}{2!}+\frac{{(x-a)^3 D^3f(a)}}{3!} + \frac{{(x-a)^4 D^4f(a)}}{4!}\cdots$$

thus,if

$$ \lim_{ n \to \infty } \int^{x}_{a} j_{n}(t){D^{n+1}}f(t){dt} =0$$

than,

$$\int^{x}_{a} {(x-t)~D^2f(t)}~dt = f(x)-f(a)-f^{\prime}(a)(x-a)  $$ 

$$ =\frac{{(x-a)^2 D^2f(a)}}{2!}+\frac{{(x-a)^3 D^3f(a)}}{3!} + \frac{{(x-a)^4 D^4f(a)}}{4!}\dots $$

$$\therefore  f(x) =  f(a)+\frac{{(x-a){D^1}f(a)}}{{1!}} + \frac{{(x-a)^2 D^2f(a)}}{2!}+\frac{{(x-a)^3 D^3f(a)}}{3!} \cdots$$

$$\therefore f(x)=\lim_{ n \to \infty } \sum^{n}_{k=0}\frac{{(x-a)^k{D^kf(a)}}}{k!}$$