How to prove this problem about the power series expanded of $C^\infty$ function? 
Problem: let $f∈C^\infty [-1,1]$, with $f^{(n)}(x)\ge0$ for all $n\in\Bbb N$ and all $x∈[-1,1]$. Show that $f$ can be expanded as a power series on $[-1,1]$

My attempt. I use the Taylor formula:
$$
f(x)=\sum_{k=0}^n \frac{f^{(k)}(0)}{k!} x^k +\frac{1}{n!} \int _{0}^{x} (x-t)^n f^{(n+1)}(t) dt. 
$$
Let
$$
R_n(x)=\frac{1}{n!} \int _{0}^{x} (x-t)^n f^{(n+1)}(t) dt.
$$
We must  prove that $R_n(x)$ tends to $0$ as $n$ tends to infinity for all $x \in[-1,1]$.
By a change of variables $x-t=u$ and $u=vx$ we get
$$
\begin{split}
|R_n(x)| & =\frac{|x|^n}{n!} \int _{0}^{1}(1-v)^n f^{(n+1)}(xv) dv \\
&\le \frac{|x|^n}{n!} \int _{0}^{1}(1-v)^n f^{(n+1)}(v) dv \\
&=|x|^{n+1}\left[f(1) - \sum_{k=0}^n \frac{f^{(k)}(0)}{k!} \right] \\
&\le |x|^{n+1} f(1).
\end{split}
$$ So we have  $R_n(x)$ tend to $0$ as $n$ tend to intfy for all $x ∈(-1,1)$. So $f$ can be expanded as a power series on $(-1,1)$.
But how to deal with the remaining points? (Indeed, I think these conditions are not sufficient to derive the remain results.)
All helps: I will thank you.
 A: As you have already shown, $f(x)$ can be expanded as a power series on the open interval $(-1,1)$. It remains to show that $f(x)$ has a power series expansion on $[-1,1]$.
Note that $f(x)$ is continuous on $[-1,1]$, so take a sequence $x_n\in (-1,1)$ such that $x_n\rightarrow 1$. By continuity, we have that $f(x_n)\rightarrow f(1)$.
This yields:
\begin{equation}
\begin{split}
f(1)=\lim_{n\rightarrow \infty}f(x_n)=\lim_{n\rightarrow \infty}\sum_{k=1}^\infty \frac{f^k(0)}{k!}{(x_n)}^k=\sum_{k=1}^\infty\frac{f^k(0)}{k!}{1}^k
\end{split}
\end{equation}
Pulling the limit inside the infinite sum is justified via the Monotone Convergence Theorem since $f^{(k)}(0)\geq 0$ by assumption and ${(x_n)}^k$ is monotonically increasing in $n$. The point $x=-1$ follows via an identical argument.
A: From the indenity $$
f(x)=\sum_{k=0}^n \frac{f^{(k)}(0)}{k!} x^k +\frac{1}{n!} \int _{0}^{x} (x-t)^n f^{(n+1)}(t) dt. 
$$
So $\sum_{k=0}^n \frac{f^{(k)}(0)}{k!} ≤f(1)$ that is to to say $\sum_{k=0}^\infty \frac{f^{(k)}(0)}{k!} $ is converges,by Abel theorem,we have $f(1)=\sum_{k=0}^\infty \frac{f^{(k)}(0)}{k!} $.
And $\sum_{k=0}^\infty \frac{f^{(k)}(0)}{k!} (-1)^k $ is also converges . By Abel theorem,$f(-1)=\sum_{k=0}^\infty \frac{f^{(k)}(0)}{k!} (-1)^k$.
