Taylor series of Infinitely differentiable function with nonnegative derivatives

Question

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 $\forall x\in(-a,a)$.

Answer 1

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.

Answer 2

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 $x \in [0,a)$.