# If $x_1,x_2,\ldots,x_n$ are the roots for $1+x+x^2+\ldots+x^n=0$, find the value of $\frac{1}{x_1-1}+\frac{1}{x_2-1}+\ldots+\frac{1}{x_n-1}$

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.

• – Arnaud D. Dec 17 '18 at 12:07

$$1$$ 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 $$\sum_{k=1}^n\frac1{x_k-1}=\lim_{z\to1}\frac{n-z-z^2-\cdots-z^n}{z^{n+1}-1} =\frac{-1-2-\cdots-n}{n+1}=-\frac n2.$$

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 $$2S=\sum_{k=1}^n\left(\frac1{x_k-1}+\frac{x_k}{1-x_k}\right) =\sum_{k=1}^n(-1)=-n.$$

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 $$\sum\limits_{i=1}^{n}\frac{1}{x_i-1}=-\frac{n}{2}$$

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 $$\sum_{r=1}^n\dfrac1{x_r-1}=\sum_{r=1}^ny_r=-\dfrac{\binom{n+1}2}{\binom{n+1}1}=?$$

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.

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

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

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

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