Pinko wrote: Let $n$ be a natural number. Find the number of real roots of the following equation: $1+\frac{x}{1}+\frac{x^2}{2}+...+\frac{x^n}{n}=0$. Looking at derivative, we get : If $n$ is even, derivative is $<0$ when $x<-1$, zero at $-1$ and $>0$ if $x>-1$ If $n$ is odd, derivative is $>0$ So If $n$ is even, expression is decreasing then increasing and minimum is reached when $x=-1$ and is $>0$, and so no real root. If $n$ is odd, expression is increasing from $-\infty\to+\infty$ and so exactly one real root.