Question about finding roots of a polynomial and studying the nature The number of real roots of the equation $1+\frac{x}{1}+\frac{x^{2}}{2}+\frac{x^{2}}{3}+\cdots+\frac{x^{7}}{7}=0$
(without factorial) is

My work
Let,$\mathrm{f}(\mathrm{x})=1+\frac{x}{1}+\frac{x^{2}}{2}+\frac{x^{3}}{3}+\cdots+\frac{x^{6}}{6}$
[Let, f has a minimum at $x=x_{0},$ where then $f^{\prime}\left(x_{0}\right)=0$]
$\Rightarrow 1+x_{0}+x_{0}^{2}+x_{0}^{3}+x_{0}^{4}+x_{0}^{5}=0$
$\Rightarrow \frac{x_{0}^{6}-1}{x_{0}-1}=0$
$\Rightarrow \frac{\left(x_{0}^{3}-1\right)\left(x_{0}^{3}+1\right)}{x_{0}-1}=0$
$\Rightarrow\left(x_{0}^{2}+x_{0}+1\right)\left(x_{0}^{2}-x_{0}+1\right)\left(x_{0}+1\right)=0$
Which has a real root $x_{0}=-1$
But, $f(-1)=1-1+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{4}-\frac{1}{5}\right)+\frac{1}{6}>0$
The $f(x)>0$ and hence $f$ has no real zeros.
Now let, $g(x)=1+\frac{x}{1}+\frac{x^{2}}{2}+\frac{x^{3}}{3}+\cdots+\frac{x^{7}}{7}$
An odd degree polynomial has at least one real root.
If our polynomial g has more than one zero, say $x_{1}, x_{2}$
Then by Role's theorem in $\left(x_{1}, x_{2}\right)$ we have $^{\prime} x_{3}$ ' such that $\mathrm{g}^{\prime}\left(x_{3}\right)=0$
$\Rightarrow 1+x_{3}+x_{3}^{2}+\cdots+x_{3}^{6}=0$
But this has no real zeros. Hence the given polynomial has exactly one real zero.
correct me if i Am wrong

Define
$$f(x)=1+\sum_{n=1}^7 \frac{x^n}{n}$$
to be the given function. Then we have
\begin{align}
f'(x)
&=\sum_{n=1}^7x^{n-1}\\
&=\cases{\frac{x^7-1}{x-1}&$x\ne1$\\7&$x=1$}\\
\end{align}
But $x^7=1$ has a single real root namely $x=1$ because it's derivative is $7x^6\ge0$. Thus $f'(x)$ has no real roots. So $f(x)$ is strictly increasing and thus has at most one real root. Using the fact that $f(x)\sim x^7/7$ for $|x|\to\infty$ we can conclude that $f(x)$ has exactly one real root (a sign change must occur).
