A polynomial of degree n with real coefficients may be represented as
$$P(x) = \sum_{i = 0}^n a_ix^i$$
which when factorised yields
$$P(x) = (x-r_1)(x-r_2)\cdots(x-r_n)$$
where $r_1, r_2,\cdots,r_n$ are the roots of the polynomial.
Now consider the first derivative of this polynomial. This will be
$$P^{'}(x) = \frac{(x-r_1)^{'}P(x)}{(x-r_1)} + \frac{(x-r_2)^{'}P(x)}{(x-r_2)} + \frac{(x-r_3)^{'}P(x)}{(x-r_3)}+ \cdots + \frac{(x-r_n)^{'}P(x)}{(x-r_n)}$$
On simplifying and rearranging we get
$$P^{'}(x) = \frac{P(x)}{(x-r_1)} + \frac{P(x)}{(x-r_2)} + \frac{P(x)}{(x-r_3)}+\cdots+ \frac{P(x)}{(x-r_n)}$$
$$P^{'}(x) = P(x)\left(\frac{1}{x-r_1} + \frac{1}{x-r_2} + \frac{1}{x-r_3} + \cdots+ \frac{1}{x-r_n}\right)$$
$$\frac{P^{'}(x)}{P(x)} = \frac{1}{x-r_1} + \frac{1}{x-r_2} + \frac{1}{x-r_3} + \cdots + \frac{1}{x-r_n}$$
for $x = 0$, we have
$$\frac{P^{'}(0)}{P(0)} = \frac{1}{-r_1} + \frac{1}{-r_2} + \frac{1}{-r_3} +\cdots + \frac{1}{-r_n}$$
$$-\frac{P^{'}(0)}{P(0)} = \sum_{i = 0}^n\frac{1}{r_i}$$
I'm an amateur at best so please share your views.
Best Answer
Seems reasonable where $P(0) \ne 0$, though a more intuitive approach exists.
If you just multiply out
$$ \prod (x-r_i) $$
The ultimate term will be the product of the roots times $(-1)^{n}$ and the penultimate will be the product of the roots lacking one root times the opposite sign times $x$.
Evaluate $P(0)$ and $P'(0)$ and take their quotient, you have your result.