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

Law of signs is covered in pre-calc/algebra. Using the standard law of signs, one has to show that there is exactly one real root between -1 and -2. Now I am rusty on how this is done (I am 60+ and I learnt it 40 years ago!) but goes something like this.
$$
\begin{align}
f(x) &= x^7/7 + x^6/x + x^5/5 + x^4/4 + x^3/3 + x^2/2 + x^1/1 + 1  && \hbox{No sign changes}\\
f(-x) &= -x^7/7 + x^6/x - x^5/5 + x^4/4 - x^3/3 + x^2/2 - x^1/1 + 1  && \hbox{7 sign changes}\\
\end{align}
$$
So we know that all the seven roots (if they exist) is to the left of $0$.
Similarly $f(x-1)$ has no sign changes but $f(x-2)$ has 7 sign changes. So all the roots have to be between $-2$ and $-1$. 
Most of the calculations are done using synthetic division and all this is covered in the precalc/algebra course. Hope someone can expand on this.
I will also dust off my precalc books and check if no one answers this
