Using Taylor series expansion as a bound I have a function $f(x)$ that has convergent Taylor series expansion around $x=0$ in the following form:
$$f(0)-xg_1(0)+\frac{x^2}{2!}g_2(0)-\frac{x^3}{3!}g_3(0)+\frac{x^4}{4!}g_4(0)-\frac{x^5}{5!}g_5(0)+\ldots$$
where $(-1)^ig_i(x)$ denotes the $i$-th derivative of $f(x)$ with respect to $x$.  I know that $g_i(x)>0$ for all $i$ at $x=0$.  I am interested in bounding $f(x)$ around small positive $x$, say $0<x\leq\epsilon$.  
I would like to make sure that, given all the facts that I described, I can make a claim that the terms $0$ through $i$ of Taylor series above form an upper bound on $f(x)$ for small positive $x$ if $i$ is even, and lower bound if $i$ is odd, or whether these facts are insufficient to make such a claim. 
(I haven't worked with Taylor series in a while and I would rather make a fool of myself in front of the experts here than at work.)
 A: Use Taylor's formula with remainder, 
$$ f(x) = f(0) + xf'(0) + \frac{x^2}{2} f^{(2)}(0) + \frac{x^3}{3!} f^{(3)}(0) + \cdots + \frac{x^m}{m!} f^{(m)}(0) + \frac{x^{m+1}}{(m+1)!} f^{(m+1)} (\xi), $$
for some $\xi \in (0,x)$. You know that $(-1)^i f^{(i)}(0) > 0$. Since $f^{(i)}$ is continuous, for some $\epsilon > 0$ it is the case that $(-1)^i f^{(i)}(\xi) > 0$ for all $\xi \in (0,\epsilon)$, and so for $x \in (0,\epsilon)$,
$$
(-1)^{(m+1)} f(x) \geq (-1)^{(m+1)} \left[f(0) + xf'(0) + \frac{x^2}{2} f^{(2)}(0) + \frac{x^3}{3!} f^{(3)}(0) + \cdots + \frac{x^m}{m!} f^{(m)}(0)\right].
$$
A: [Edit: I think there's a fatal flaw in this proof, as Didier Piau pointed out: it only works for alternating series where the terms are monotonically decreasing in absolute value.]
Abstracting a bit, you have a function with a convergent series representation near $x=0$:
$$
f(x) = c_0 - c_1 x + c_2 x^2 - c_3 x^3 + c_4 x^4 \cdots,
$$
where the $c_j$ are all positive. (Your problem didn't specify that $f(0)$ is positive, but shifting $f$ by a constant doesn't change the problem.) It's true that the $c_j$ are related to the derivatives of $f$ at $x=0$ (although I think you're missing some factorials), but that's not important for what follows.
You ask, for $x$ small and positive, whether the truncations of the infinite series form alternately lower and upper bounds for $f(x)$. The answer is yes, simply because it's a convergent alternating series: in any convergent alternating series, the truncations alternate between lower and upper bounds. (Proof: draw the picture!)
