Prove $\sum_{r=1}^n\frac{a^s_r}{f'(a_r)} = 0$ Show that for $f(x) = (x-a_1)(x-a_2)...(x-a_n)$, where $a_1>a_2>...>a_n$,
$$\sum_{r=1}^n\frac{a^s_r}{f'(a_r)} = 0$$
for $s = 0, 1 ,2, ..., n-2$.
I was told I need to find to prove this first
$$\frac{x^{n-1}}{f(x)}=\sum_{r=1}^n\frac{\frac{a^{n-1}_r}{f'(a_r)}}{x-a_r}
$$
and use it to solve the question. But I don't know how to even start with proving either of those. I'd really appreciate help for this.
 A: Let
$$F(x)=x^{n-1}-\sum_{r=1}^n\frac{a^{n-1}_r}{f'(a_r)}\frac{f(x)}{x-a_r}
$$
Note that the degree of $F(x)$ is a polynomial with degree $n-1$ and that
$$ \frac{f(x)}{x-a_r}\bigg|_{x=a_j}=0 \text{ if }r\neq j, \frac{f(x)}{x-a_r}\bigg|_{x=a_r}=f'(a_r). $$
Now 
$$ F(a_r)=0, r=1,2,\cdots,n $$
which means that $F(x)$ has $n$ different roots. Hence one must have $F(x)\equiv0$. So
$$x^{n-1}=\sum_{r=1}^n\frac{a^{n-1}_r}{f'(a_r)}\frac{f(x)}{x-a_r} $$
or
$$\frac{x^{n-1}}{f(x)}=\sum_{r=1}^n\frac{\frac{a^{n-1}_r}{f'(a_r)}}{x-a_r}.
$$
Letting $x=0$ gives the identity for $s=n-2$. The rest should be easy to handle and I omit the detail.
A: Good proof xpaul except a typo in the first equation for $F(x)$
So we get $$\frac{x^{n-1}}{f(x)}=x^{n-1}=\sum_{r=1}^n\frac{a^{n-1}_r}{f'(a_r)}\frac{1}{x-a_r}$$
And we can get, when $x=0$,   $\sum_{r=1}^n\frac{a^{n-2}_r}{f'(a_r)}=0$
Differentiate the first equation, $$\frac{(n-1)x^{n-2}f(x)-x^{n-1}f'(x)}{f^2(x)}=-\sum_{r=1}^n\frac{a^{n-1}_r}{f'(a_r)}\frac{1}{(x-a_r)^2}$$
similarly, let x=0, we get $\sum_{r=1}^n\frac{a^{n-3}_r}{f'(a_r)}=0$
This can continue till (n-1)th derivative. 
A: Let $R$ be some positive number greater than $\max(|a_1|,|a_n|)$. We have that
$$ f_s(R) = \oint_{\|z\|=R}\frac{z^s}{f(z)}\,dz $$
by Cauchy's theorem, does not depend on $R$ but just on a sum of residues. On the other hand, if $s\leq \deg f(z)-2$ we have that $\left|\,f_s(R)\right|\ll \frac{1}{R}$ for large $R$, hence $f_s(R)\equiv 0$. By expressing $f_s(R)$ in terms of residues,
$$ f_s(R) = 2\pi i\sum_{k=1}^{n}\text{Res}\left(\frac{z^s}{f(z)},z=a_k\right) $$
and since $a_1,\ldots,a_k$ are simple poles for $\frac{z^s}{f(z)}$,
$$ \text{Res}\left(\frac{z^s}{f(z)},z=a_k\right) = \lim_{z\to a_k}\frac{z^s(z-a_k)}{f(z)}\stackrel{dH}{=}\lim_{z\to a_k}\frac{s z^{s-1}(z-a_k)+z^s}{f'(z)} = \frac{a_k^s}{f'(a_k)}.$$
The claim easily follows.
