Taylor series of Infinitely differentiable function with nonnegative derivatives Let $f(x)$ be a nonnegative and infinitely differentiable function on $[-a,a]$ to $\mathbb{R}$ such that $\forall x\in[-a,a]:f^{(n)}(x)\ge0$.
Prove that the series:
$$\sum_{i=1}^\infty \frac{f^{(i)}(0)}{i!}x^i$$
converges $\forall x\in(-a,a)$.
 A: First, recall that
$$
f(x)-\sum_{n=0}^m\frac{f^{(n)}(0)}{n!}x^n=\frac{x^{m+1}}{m!}\int_0^1(1-t)^m f^{(m+1)}(tx)dt
$$
which can be proved easily using integration by parts.
So, for $0\leq x<a$ we have 
$$
\forall\, m\geq 0,\quad\sum_{n=0}^m\frac{f^{(n)}(0)}{n!}x^n\leq f(x)
$$
This proves the convergence of the considered series because it has non-negative terms and bounded partial sums.
Now, this can be used to prove the absolute convergence of the series when $x\in(-a,0)$. Indeed, for $x\in(-a,0)$ we have
$$
 \sum_{n=0}^\infty\left\vert\frac{f^{(n)}(0)}{n!}x^n\right\vert
= \sum_{n=0}^\infty\frac{f^{(n)}(0)}{n!}\left\vert x\right\vert^n\leq f(\vert x\vert) 
$$
So the series converges also in this case.
A: First observe that it suffices to prove the claim for $x \in [0,a)$ (why?).
For $n \in \Bbb{N}$ define
$$
f_n(x) := \sum_{i=1}^n \frac{f^{(i)} (0)}{i!} \cdot x^i.
$$
We then have
$$
f_n^{(n+1)}(x) = 0 \leq f^{(n+1)}(x)
$$
for all $x \in [0, a)$ and also $f_n^{(i)} (0) = f^{(i)} (0)$ for $0 \leq i \leq n$.
Using this, we inductively see
$$
f_n^{(n+1-i)}(x) \leq f^{(n+1-i)} (x) \, \forall x \in [0,a)
$$
for $0 \leq i \leq n+1$ (integrate the previous inequality).
For $i=n+1$, this implies $f_n (x) \leq f(x)$ for $x \in [0,a)$.
It remains to observe that the sequence $(f_n(x))_n$ is increasing and bounded for each $x \in [0,a)$.
