Real roots of the equation $1+\sum_{r=1}^{7}\frac{x^{r}}{r} = 0$ The number of real roots of the equation $\displaystyle 1+\frac{x}{1}+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+\frac{x^5}{5}+\frac{x^6}{6}+\frac{x^7}{7} = 0$
$\bf{My\; Try}::$ Let $\displaystyle f(x) = 1+\frac{x}{1}+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+\frac{x^5}{5}+\frac{x^6}{6}+\frac{x^7}{7}$
Now $\displaystyle f^{'}(x) = 1+x+x^2+x^3+x^4+x^5+x^6$
and $\displaystyle f^{''}(x) = 1+2x+3x^2+4x^3+5x^4+6x^5$
$\displaystyle f^{'''}(x) = 0+2+6x+12x^2+20x^3+30x^4 = 2\left(1+3x+6x^2+10x^3+15x^4\right)$
Now I did not understand how can i solve it
Help Required
Thanks

Note that
$$f'(x)=\sum_{k=0}^6 x^k=\frac{x^7-1}{x-1}.$$
Now, it is not too hard to see that $f'(x)>0$ for all $x\in\mathbb{R}$ (consider regions $x<-1$, $-1<x<0$, $0<x<1$ and $x>1$). So, there is a real root because the degree of the polynomial is odd, but there is only one because the function is monotonically increasing.
