What is the value of $$\sum _{k=1}^n\frac{x_k}{x_k-1}$$ given that $x_1,x_2,\dots,x_n $ are the roots of the equation $x^n-3x^{n-1}+2x+1=0\,$?

I wrote it as $ n+\sum _{k=1}^n\frac{1}{x_k-1}$ but didn't really help much.

I think Vieta's formulas come in handy here, but I didn't get to the part where I could actually use any.

Could I have some hints on how to get this done?

Thank you. The answer is $S=3n-5$.

  • 3
    $\begingroup$ What do you want to prove? $\endgroup$ – Marios Gretsas Aug 20 '17 at 7:52
  • $\begingroup$ Just edited, check it out again please @MariosGretsas $\endgroup$ – Alexander Aug 20 '17 at 7:53
  • $\begingroup$ Already tried it, but I didn't really get to the point where I can use the fact that x1+x2+...+xn=3 or x1*x2*...*xn=(-1)^n. @rtybase $\endgroup$ – Alexander Aug 20 '17 at 8:22
  • $\begingroup$ @Alexander it turned to be a simple application of derivative. $\endgroup$ – rtybase Aug 20 '17 at 9:03
  • $\begingroup$ ...where $x_i$ denotes what, exactly? $\endgroup$ – John Hughes Mar 26 at 19:19

Hint. You are on the right track. Let $x_k-1=1/y_k$ then $$\sum _{k=1}^n\frac{x_k}{x_k-1}=n+\sum _{k=1}^n y_k.$$ For $x=1+1/y$ we have that \begin{align*}0&=y^n(x^n-3x^{n-1}+2x+1)= y^n((1/y+1)^n-3(1/y+1)^{n-1}+2(1/y+1)+1)\\ &=(y+1)^n-3y(y+1)^{n-1}+2(y^n+y^{n-1})+y^n\\ &=(1-3+2+1)y^n+(n-3(n-1)+2)y^{n-1}+o(y^{n-1})\\ &=y^{n}-(2n-5)y^{n-1}+o(y^{n-1}). \end{align*} Hence the sum of the roots, i.e. $\sum _{k=1}^n y_k$, is equal to $2n-5$ and we finally get $$\sum _{k=1}^n\frac{x_k}{x_k-1}=n+(2n-5)=3n-5.$$

  • $\begingroup$ Could you please expend on where you got the n's and numbers from in the second to last line? $\endgroup$ – Alexander Aug 20 '17 at 8:46
  • $\begingroup$ @Alexander Use binomial theorem $(y+1)^N=y^N+Ny^{N-1}+o(y^{N-1})$. $\endgroup$ – Robert Z Aug 20 '17 at 8:55

We have $$P(x)= x^n-3x^{n-1}+2x+1=\prod\limits_{k=1}^{n}\left(x-x_k\right)$$ From which $$P'(x)=nx^{n-1}-(n-1)3x^{n-2}+2=\sum\limits_{i=1}^{n}\prod\limits_{k=1,k\ne i}^{n}\left(x-x_k\right)$$ then $$\frac{P'(x)}{P(x)}=\sum\limits_{i=1}^{n}\frac{1}{x-x_i}$$

But $$S=n+\sum\limits_{k=1}^{n}\frac{1}{x_k-1}=n-\frac{P'(1)}{P(1)}=n-(n-3(n-1)+2)=3n-5$$


Let $p(x)=x^{n}-3x^{n-1}+2x+1=\prod_{i=1}^{n}(x-x_i)$. Then $$\frac{p'(x)}{p(x)}=\frac{\sum_{i=1}^{n}\prod_{1\leq j\leq n, j\neq i}(x-x_i)}{\prod_{i=1}^{n}(x-x_i)}=\sum_{i=1}^{n}\frac{1}{x-x_i}$$

Therefore $$\frac{xp'(x)}{p(x)}=\sum_{i=1}^{n}\frac{x}{x-x_i}=\sum_{i=1}^{n}\left(1+\frac{x_i}{x-x_i}\right)=n+\sum_{i=1}^{n}\frac{x_i}{x-x_i}$$

Evaluate at $x=1$, which we can because $p(1)=1\neq0$, to get


  • $\begingroup$ It's a bit more efficient to use $\frac{p'(x)}{p(x)}=\frac{d\log(p(x))}{dx}$ and $p(x)=\prod_{i=1}^n (x-x_i)$ so that $\log\left(p(x)\right)=\sum_{i=1}^n \log(x-x_i)$. $\endgroup$ – Mark Viola Mar 26 at 19:49
  • $\begingroup$ @MarkViola It only replaces one equality by another equality. There is not gain. $\endgroup$ – user647486 Mar 26 at 19:50
  • 2
    $\begingroup$ @MarkViola Moreover, this way the proof is exclusively algebraic and therefore can be used in fields other than the complex numbers. $\endgroup$ – user647486 Mar 26 at 20:02
  • $\begingroup$ I believe that there is significant "gain" in fluidity and ease. I left the comment as a supplement for future readers. $\endgroup$ – Mark Viola Mar 26 at 22:50

We have for a polynomial $P(x)$ of degree $n$ with roots $\alpha_1, \alpha_2, \ldots \alpha_n$,

$\frac{P'(a)}{P(a)} = \frac{1}{a-\alpha_1}+\frac{1}{a-\alpha_1}+\cdots \frac{1}{a-\alpha_n}$

Hence $\sum _{k=1}^n\frac{1}{x_k-1} = -\frac{P'(1)}{P(1)} = 2n-5$ and hence the sum of the given series is $3n-5$

  • $\begingroup$ Right, I was slower with my answer ... (+1) though! $\endgroup$ – rtybase Aug 20 '17 at 9:01
  • $\begingroup$ @rtybase: thank you for the upvote, I have returned the favour :) $\endgroup$ – Hari Shankar Aug 20 '17 at 9:04

I think the $x_k$'s are the roots of the polynomial, right?

Anyway, put $y_{k} = \frac{x_k}{x_k - 1}$, so that $x_k = \frac{y_k}{y_k - 1}$. Since $x_k$'s are the roots of the polynomial $f(x) = x^n - 3x^{n-1} + 2x + 1$, we have $f(y_k/(y_k-1)) =0$ for all $k$. Using this, one can find the polynomial that has $y_1, \dots, y_k$ as roots, and Vieta's formula will give the answer.

  • $\begingroup$ I don't understand why f(..)=0 $\endgroup$ – Gaboru Mar 26 at 19:34
  • $\begingroup$ @Gaboru Because $f(x_k) =0$ and $x_k = y_k/(y_k - 1)$. $\endgroup$ – Seewoo Lee Mar 26 at 19:34

One approach is to put $y=x-1$ so the sum becomes terms of the form $\cfrac {y+1}y=1+\cfrac 1y$

Now $y$ satisfies $$(y+1)^n-3(y+1)^{n-1}+2(y+1)+1=0$$

If we now divide through by $y^n$ and set $z=\cfrac 1y$ (it is possible to omit this step if you know what you are doing, and read off the relevant data from the polynomial for $y$) we get $$(1+z)^n-3z(1+z)^{n-1}+2z^{n-1}(1+z)+z^n=0$$and you should have a form you can use to conclude.


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