Suppose $f\in C^{\infty}[a,b]$, and for any $n\geq 1$, $f^{(n)}(x)$ is non-negative. Prove that $$f(x)=\sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!} (x-a)^n\,,\, \forall x\in [a,b]\,. $$ Hint: using $R_n(x)=\frac{1}{n!}\int_a^x f^{(n+1)}(t) (t-a)^n dt $ and $f^{(n)}(x)$ is increasing.
I have tried in the following way: $$\begin{aligned} R_n(x)&=\frac{1}{n!}\int_a^x f^{(n+1)}(t) (t-a)^n dt \\&= \frac{1}{n!} f^{(n+1)}(\xi) \int_a^x (t-a)^n dt\\&\leq \frac{f^{(n+1)}(b)}{(n+1)!} (b-a)^{n+1}. \end{aligned}$$ But I don’t know how to prove last step, which is $\frac{f^{(n+1)}(b)}{(n+1)!} (b-a)^{n+1} \to 0$ as $n\to\infty$.
Or there is another method to solve this problem, looking forward to your answers! Thank you!
Since fⁿ⁺¹(b) constant, if (b-a)$\le$ 1 then (b-a)ⁿ⁺¹/(n+1)! goes to 0 as n tend to ∞ automatically.
But if (b-a)>1 then take aₙ=(b-a)ⁿ⁺¹/(n+1)!
Now since $\lim_{n\to ∞}$ aₙ₊₁/aₙ=$\lim_{n\to ∞}$ (b-a)/(n+2)=0 < 1,( the property : {aₙ} be a sequence of positive real numbers then if $\lim_{n\to ∞}$ aₙ₊₁/aₙ<1 then $\lim_{n\to ∞}$ aₙ = 0 and if $\lim_{n\to ∞}$ aₙ₊₁/aₙ>1 then $\lim_{n\to ∞}$ aₙ tends to ∞).
$\lim_{n\to ∞}$ aₙ = 0 , as required.