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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.

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If you are allowed to use the fact that the full series has sum $e^{-x}$, you are finished, for already $e^{-x}-1+x$ is clearly positive when $x\ge 1$, indeed somewhat earlier. The challenge would be to prove it without the $e^{-x}$ result. – André Nicolas Feb 10 '12 at 1:30
what if it isn't allowed to use this property? What is the approach then? – user19821 Feb 10 '12 at 3:10
Sorry, I was away. Since then, that version of the question has been fully answered by Didier Piau. – André Nicolas Feb 10 '12 at 7:13
up vote 4 down vote accepted

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 $x$.

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great, thanks Didier – user19821 Feb 10 '12 at 22:01

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.

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