How to show $1 +x + x^2/2! + \dots+ x^{2n}/(2n)!$ is positive for $x\in\Bbb{R}$? 
How to show $1 + x + \frac{x^2}{2!} + \dots+ \frac{x^{2n}}{(2n)!}$ is positive for $x\in\Bbb{R}$?

I realize that it's a part of the Taylor Series expansion of $e^x$ but can't proceed with this knowledge? Also, I can't figure out the significance of $2n$ being the highest power.
 A: By induction, assume that 
$$f_{2n}(x)=1+x+\frac{x^2}2+\cdots\frac{x^{2n}}{(2n)!}>0.$$
Then the antiderivative 
$$f_{2n+1}(x)=1+x+\frac{x^2}2+\cdots\frac{x^{2n+1}}{(2n+1)!}$$ is a growing function and has a single root.
Then the antiderivative 
$$f_{2n+2}(x)=1+x+\frac{x^2}2+\cdots\frac{x^{2n+2}}{(2n+2)!}$$ has a single minimum, which occurs at this root, let $r$.
The value at the minimum is easy to compute as
$$f_{2n+2}(r)=f_{2n+1}(r)+\frac{r^{2n+2}}{(2n+2)!}=0+\frac{r^{2n+2}}{(2n+2)!},$$
which is a positive number.
A: If we use Taylor's theorem with an integral remainder we get:
$$ 1+x+\frac{x}{2}+\ldots+\frac{x^{2n}}{(2n)!} = e^x-\frac{1}{(2n)!}\int_{0}^{x}t^{2n}e^{x-t}\,dt \tag{1}$$
hence in order to prove the non-negativity of the LHS it is enough to show that:
$$ \int_{0}^{x} t^{2n}e^{-t}\,dt \leq (2n)! \tag{2}$$
holds for any $x\in\mathbb{R}$. $(2)$ is trivial if $x\leq 0$, and if we assume $x>0$, since the integrand function is non-negative, we get:
$$ \int_{0}^{x}t^{2n}e^{-t}\,dt \leq \int_{0}^{+\infty}t^{2n}e^{-t}\,dt = \Gamma(2n+1)=(2n)!,\tag{3}$$
proving our claim.
A: Let $P_n(x)$ your polynomial. The Taylor-Lagrange formula gives for $x\in \mathbb{R}$ that there exists $c$ such that $\exp(x)=P_n(x)+\frac{x^{2n+1}}{(2n+1)!}\exp(c)$ for some $c$ depending on $x$. Now suppose that there exists $x\in \mathbb{R}$ such that $P_n(x)=0$. Obviously, we have $x<0$. Then $\exp(x)=\frac{x^{2n+1}}{(2n+1)!}\exp(c)$, and $\exp(x-c)=\frac{x^{2n+1}}{(2n+1)!}<0$, a contradiction. Hence $P_n(x)$ is never zero, and as a continuous function, it has a constant sign. As $P_n(0)=1$, we are done. 
