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Show that the polynomial $p(x)=1-x+\frac{x^2}{2}-\frac{x^3}{3}+\cdots+(-1)^n\frac{x^n}{n}$ has exactly one real zero if $n$ is odd, but none if $n$ is even.

If $n$ is odd $\displaystyle\lim_{x \to - \infty}p(x)=+\infty$, $\displaystyle\lim_{x \to + \infty}p(x)=-\infty$ and $p'(x)<0$ for all $x \in \mathbb{R}$, then by the Intermediate Value Theorem, follows that there is only real value $c$ such that $p(c)=0$.

But, if $n$ is even I know not as do it. Someone help me. Thanks!

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    Hint: What is $(1+x)p^\prime(x)$? – 2017-01-15
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    @G.Sassatelli I think he meant IVT with $p^\prime(x)<0$. – 2017-01-15
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    @LeonSot Sorry, I missed the condition. – 2017-01-15
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    But in this case ($n$ odd) $p'(x)<0$, then $p$ is strictly decreasing. Because this there is only one $c \in \mathbb{R}$ such that $p(c)=0$. – 2017-01-15

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