My high school teacher taught me a easy way to calculate the
$$\sum\frac{1}{a^n}\qquad n\in \mathbb{Z}$$ where $a$ is the roots of a quadratic equation in the form $f(x)=x^2+ ax +b$.
He told me to find the derivative of $f(x)$ $f'(x) = 2x + a$.
The find $-f'(x)/f(x)$ which is going to be an infinite series.
Suppose $\frac{-f'(x)}{f(x)} = a_0 + a_1x + a_2x^2 + a_3x^3.....$
He stated that:
$\sum\frac{1}{a^0}= a_0$
$\sum\frac{1}{a^1}= a_1$
$\sum\frac{1}{a^2}= a_2$ And so on...
My doubt is why is this true? I wanted a proof.
PS: I hope my MathJax formatting is correct. I am new to this community so I would like recommendations from members.
Let $f(x)$ be a monic polynomial of degree $d$ with all roots $\alpha_1,\ldots, \alpha_d$ simple. In other words, $$ f(x)=(x-\alpha_1)(x-\alpha_2)\cdots (x-\alpha_d)=\prod_{k=1}^d(x-\alpha_k).$$ Then $$\ln f(z)=\sum_{k=1}^d\ln(z-\alpha_k) $$ and with $g(z):=-\frac{f'(z)}{f(z)}$, we observe $$ -g(z)=\frac{f'(z)}{f(z)}=\frac{\mathrm d}{\mathrm dz}\ln f(z)=\sum_{k=1}^d\frac{\mathrm d}{\mathrm dz}\ln(z-\alpha_k)=\sum_{k=1}^d\frac1{z-\alpha_k}.$$ Now use the geometric series $$\frac1{\alpha_k-z}=\sum_{n=0}^\infty \alpha_k^{n+1}z^n$$ so that by comparing coefficients of equal powers of $z$ in the power series, we conclude $$a_n= \sum_{k=1}^ d\alpha_k^{n+1}.$$