How to show $\sum_{k \geq 2} \frac{(-x)^k}{k!} \geq 0$ for large x It's probably a very silly question. 
I could only show though that
$$
\sum_{k \geq 2} \frac{(-x)^{k}}{k!} = \sum_{k \geq 0} \frac{(-x)^{k}}{k!} -1 +x= e^{-x}-1+x
$$ 
which tends to infinity as $x \to \infty$. And also taking the ration of consecutive terms 
$$
\frac{a_{k+1}}{a_{k}} = -\frac{x}{k+1}
$$
for positive x, so the sum is dominated by the first term, which is positive, $\frac{x^2}{2!}$. 
I'm not completely sure of these derivations, so it would be good if someone could help.
 A: Consider the function $u:x\mapsto\mathrm e^{-x}-1+x$. Its derivative is $u':x\mapsto-\mathrm e^{-x}+1$. Since $\mathrm e^0=1$ and the exponential is an increasing function, $u'(x)\lt u'(0)=0$ if $x\lt0$ and $u'(x)\gt u'(0)=0$ if $x\gt0$. Thus, for every $x$, $u(x)\geqslant u(0)=0$.
This proves that $\sum\limits_{k=2}^{+\infty}\frac{(-x)^k}{k!}\geqslant0$ for every real number $x$, and  that $\sum\limits_{k=2}^{+\infty}\frac{(-x)^k}{k!}\gt0$ for every $x\ne0$.

Edit: In the (somewhat odd) situation of not knowing that the sum of the series $s(x)=\sum\limits_{k=2}^{+\infty}\frac{(-x)^k}{k!}$ is $\mathrm e^{-x}-1+x$ but being allowed to differentiate $s(x)$ term by term, one can proceed as follows. First,
$$
s'(x)=\sum\limits_{k=2}^{+\infty}(-1)^kk\frac{x^{k-1}}{k!}=\sum\limits_{k=1}^{+\infty}(-1)^{k+1}\frac{x^{k}}{k!}=x-s(x).
$$
Hence the function $t:x\mapsto s(x)\mathrm e^x$ is such that $t'(x)=(s'(x)+s(x))\mathrm e^x=x\mathrm e^x$. In particular, $t'(x)\lt0$ if $x\lt0$ and $t'(x)\gt0$ if $x\gt0$. Since $t(0)=0$ this yields $t(x)\gt0$ for every $x$, and finally, $s(x)=\mathrm e^{-x}t(x)\gt0$ for every $x$.
A: If $f(x) \to +\infty$ as $x \to \infty$, then certainly $f(x) \ge 0$ for sufficiently large $x$.  It's not dominated by the first term: e.g. the third term $x^4/4!$ is greater than the first term when $x$ is large.
