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Let $x_1,x_2,\ldots,x_n$ be the roots for $1+x+x^2+\ldots+x^n=0$. Find the value of $$P(1)=\frac{1}{x_1-1}+\frac{1}{x_2-1}+\ldots+\frac{1}{x_n-1}$$

Source: IME entrance exam (Military Institute of Engineering, Brazil), date not provided (possibly from the 1950s)

My attempt: Developing expression $P(1)$, replacing the 1 by $x$, follows $$P(x)=\frac{(x_2-x)\cdots (x_n-x)+\ldots+(x_1-x)\cdots (x_{n-1}-x)}{(x_1-x)(x_2-x)\cdots (x_n-x)}$$

As $x_1,x_2,\ldots,x_n$ are the roots, it must be true that

$$Q(x)=(x-x_1)\cdots(x-x_n)=1+x+x^2+\ldots+x^n$$ and $$Q(1)=(1-x_1)\cdots(1-x_n)=n+1$$ Therefore the denominator of $P(1)$ is $$(-1)^{n} (n+1).$$ But I could not find a way to simplify the numerator.

Another fact that is probably useful is that $$1+x^{n+1}=(1-x)(x^n+x^{n-1}+\ldots+x+1)$$ with roots that are 1 in addition of the given roots $x_1,x_2,\ldots,x_n$ for the original equation, that is

$$x_k=\text{cis}(\frac{2k\pi}{n+1}),\ \ k=1,\ldots,n.$$

This is as far as I could go...

Hints and full answers are welcomed.

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$1$ and $x_1,\ldots,x_n$ are the roots of $z^{n+1}-1=0$. They are the $(n+1)$-th roots of unity. Then $1/(z^{n+1}-1)$ has a partial fraction expansion of the form $$\frac1{z^{n+1}-1}=\frac a{z-1}+\sum_{k=1}^n\frac{b_k}{x_kz-1}.$$ Multiplying by $z-1$ and setting $z=1$ gives $a=1/(n+1)$. Multiplying by $x_kz-1$ and setting $z=1/x_k$ gives $b_k=1/(n+1)$. Therefore $$\sum_{k=1}^n\frac1{x_kz-1}=\frac{n+1}{z^{n+1}-1}-\frac1{z-1}.$$ Letting $z\to1$ gives $$\sum_{k=1}^n\frac1{x_k-1}=\lim_{z\to1}\frac{n-z-z^2-\cdots-z^n}{z^{n+1}-1} =\frac{-1-2-\cdots-n}{n+1}=-\frac n2.$$

ADDED IN EDIT.

Here's a trick proof. Let $S$ denote the sum in question. As the reciprocals of the $x_k$ are the $x_k$ again (in a different order) then $$S=\sum_{k=1}^n\frac1{x_k^{-1}-1}=\sum_{k=1}^n\frac{x_k}{1-x_k} $$ and $$2S=\sum_{k=1}^n\left(\frac1{x_k-1}+\frac{x_k}{1-x_k}\right) =\sum_{k=1}^n(-1)=-n.$$

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Applying the same technique as per this answer $$Q(x)=\prod\limits_{k=1}^n(x-x_k) \text{ and } Q'(x)=\sum\limits_{i=1}^{n}\prod\limits_{k=1,k\ne i}^{n}\left(x-x_k\right)$$ then $$\frac{Q'(x)}{Q(x)}=\sum\limits_{i=1}^{n}\frac{1}{x-x_i}$$ or $$-\frac{Q'(1)}{Q(1)}=\sum\limits_{i=1}^{n}\frac{1}{x_i-1}$$ But $Q(1)=n+1$ and $Q'(x)=1+2x+3x^3+...+nx^{n-1}\Rightarrow Q'(1)=\frac{n(n+1)}{2}$ and $$\sum\limits_{i=1}^{n}\frac{1}{x_i-1}=-\frac{n}{2}$$

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We have $$x^{n+1}-1=0\ \ \ \ (1)$$ with $x\ne1$

Set $\dfrac1{x-1}=y\iff x=\dfrac{y+1}y$

Replace the value of $x$ in terms of $y$ in $(1)$ to form an $n$ degree equation in $y$ $$\left(\dfrac{y+1}y\right)^{n+1}=1$$

$$\implies\binom{n+1}1 y^n+\binom{n+1}2 y^{n-1}+\cdots+1=0$$

Now apply Vieta's formula to find $$\sum_{r=1}^n\dfrac1{x_r-1}=\sum_{r=1}^ny_r=-\dfrac{\binom{n+1}2}{\binom{n+1}1}=?$$

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Simply note that $1,x_1,\ldots, x_n$ are roots of $$ x^{n+1} -1 = 0.$$ Thus we have $0,x_1-1,\ldots, x_n-1$ are roots of $$ (x+1)^{n+1} -1 = x\sum_{j=0}^{n} \binom{n+1}{j+1}x^j = 0. $$ Hence, $z_i = x_i -1$, $1\leq i \leq n$ are roots of $$ \sum_{j=0}^{n} \binom{n+1}{j+1}z^j = 0, $$ and by Vieta's formula, $$ \sum_{i=1}^n \frac{1}{z_i} = \frac{\sum_{i=1}^n z_1z_2\cdots z_{i-1}\widehat{z_i}z_{i+1}\cdots z_n}{z_1z_2\cdots z_{n-1}z_n} = -\frac{\binom{n+1}{2}}{\binom{n+1}{1}} = -\frac{n}{2}, $$ where $\widehat{z_i}$ means the variable is omitted in the calculation.

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First, the denominator is, I think, $(-1)^nQ(1)$. Here is a hint for the numerator: how can you write $Q’(x)$?

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We solve it using complex residues by way of enrichment. With the function

$$f(z) = \frac{1}{z-1} \frac{(n+1)/z}{z^{n+1}-1}$$

we have for $\zeta_k = \exp(2\pi i k/(n+1))$ with $1\le k\le n$ that

$$\mathrm{Res}_{z=\zeta_k} f(z) = \frac{1}{z-1} \left.\frac{(n+1)/z}{(n+1)z^n}\right|_{z=\zeta_k} = \frac{1}{\zeta_k-1}.$$

Residues sum to zero so we have

$$\sum_{k=1}^{n} \frac{1}{\zeta_k-1} + \mathrm{Res}_{z=0} f(z) + \mathrm{Res}_{z=1} f(z) + \mathrm{Res}_{z=\infty} f(z) = 0.$$

The residue at infinity is zero by inspection. The residue at $z=0$ is $n+1$, also by inspection. We have for the residue at $z=1$

$$\mathrm{Res}_{z=1} f(z) = \mathrm{Res}_{z=1} \frac{1}{(z-1)^2} \frac{(n+1)/z}{1+z+\cdots +z^{n}} \\ = \left.\left( \frac{(n+1)/z}{1+z+\cdots +z^{n}} \right)'\right|_{z=1} \\ = \left.\left( - \frac{(n+1)/z^2}{1+z+\cdots +z^{n}} - \frac{(n+1)/z \times (1+2z+\cdots+nz^{n-1})} {(1+z+\cdots +z^{n})^2} \right)\right|_{z=1} \\ = -1 - \frac{(n+1) \times \frac{1}{2} n (n+1)}{(n+1)^2}.$$

Now with

$$\sum_{k=1}^{n} \frac{1}{\zeta_k-1} = - \mathrm{Res}_{z=0} f(z) - \mathrm{Res}_{z=1} f(z)$$

we obtain

$$-(n+1) + 1 + \frac{1}{2} n.$$

and therefore our answer is

$$\bbox[5px,border:2px solid #00A000]{ \sum_{k=1}^{n} \frac{1}{\zeta_{k}-1} = - \frac{n}{2}.}$$

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We notice that the equation: $$x^{n} + x^{n-1} + ... +x +1 = 0 $$ is a geometric sequence and can be rewritten as: $$x^{n} + x^{n-1} + ... +x +1 = \sum_{i=0}^{n}x^n = \frac{1-x^{n+1}}{1-x} = 0 $$ From the denominator $x-1 $ we conclude that $x\ne1$, but we can rewrite $ 1 = e^{2\pi k} $. So from: $$ 1-x^{n+1} = 0 \rightarrow x^{n+1} = e^{2\pi k} $$ $$ x_k = e^{2\pi k/(n+1)} $$ Where $k \ne 0$ and goes from $1$ to $n$. Because $k\ne 0$ then $x\ne1$ and there is no problem with the denominator anymore. The sum becomes: $$ \sum_{k=1}^{n} \frac{1}{1-x_k} = \sum_{k=1}^{n} \frac{1}{1-e^{2\pi k/(n+1)}} = \sum_{k=1}^{n} \frac{e^{-\pi k/(n+1)}}{e^{-\pi k/(n+1)}-e^{\pi k/(n+1)}} = \sum_{k=1}^{n} \frac{\cos(\pi k/(n+1))-i\sin(\pi k/(n+1))}{-2i\sin(\pi k/(n+1))} = $$ $$ \sum_{k=1}^{n} \frac{1}{2} + i\frac{\cot(\pi k/(n+1))}{2} = \frac{n}{2} + \frac{i}{2}\sum_{k=1}^{n}\cot(\pi k/(n+1)) = \frac{n}{2} $$ From Wolfram alpha $ \sum_{k=1}^{n}\cot(\pi k/(n+1)) = 0 $ but you can prove it easily. An example: $$ \cot(\frac{n\pi}{n+1}) = \cot(\frac{(n+1)\pi-\pi}{n+1}) = \cot(\pi- \frac{\pi}{n+1}) = -\cot(\frac{\pi}{n+1}) $$ which will cancel out with the first term and so on with other terms.

NOTE: I conducted the calculations for the sum $ \sum_{k=1}^{n} \frac{1}{1-x_k} $.

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