# Series involving complex roots: $\frac{1}{2-a_1} + \frac{1}{2-a_2} + \dots + \frac{1}{2-a_{n-1}} = \frac{(n-2)2^{n-1}+1}{2^n - 1}$

$$\frac{1}{2-a_1} + \frac{1}{2-a_2} + \dots + \frac{1}{2-a_{n-1}} = \frac{(n-2)2^{n-1}+1}{2^n - 1}$$

Here $1,a_1,a_2,\dots,a_{n-1}$ are $n$-th roots of unity

I know the sum of roots is 0. I think the series should be some sort of telescopic.

• I am new here and i dont know how to type it – thebeatle Jun 26 '15 at 16:13
• Hint: $$\sum_{k=1}^n\frac{1}{z - \lambda_k} = \frac{P'(z)}{P(z)} \quad\text{ where }\quad P(z) = \prod_{k=1}^n (z - \lambda_k)$$ Now choose a right $P(z)$. – achille hui Jun 26 '15 at 16:25
• I chose p(z) as 2^n-1 and p'(z) turns out to be bunch of other factors. – thebeatle Jun 26 '15 at 16:31
• The $p(z)$ you've chosen is a constant. Small hint: the series on the left is missing one root of the unity. Add it to the both sides – uranix Jun 26 '15 at 16:32
• The LHS of your equation isn't exactly $\sum_{k=1}^n \frac{1}{z-\lambda_k}$, it is missing the term for $1$. If you subtract $\left.\frac{P'(z)}{P(z)}\right|_{z=2}$ by $\frac{1}{2-1}$ and simplify, you will get your RHS. BTW, if $P(z) = z^n - 1$, $P'(z)$ will be simply $n z^{n-1}$. – achille hui Jun 26 '15 at 16:33

Let $y=\dfrac1{2-x}\iff x=\dfrac{2y-1}y$ where $x^n=1$
$\implies\left(\dfrac{2y-1}y\right)^n=1$
$\displaystyle\iff y^n(2^n-1)-\binom n1 2^{n-1}y^{n-1}+\cdots+(-1)^n=0$
Using Vieta's formula, $$\displaystyle\sum_{r=0}^{n-1}y_r=\dfrac{\binom n1 2^{n-1}}{2^n-1}=\dfrac{n 2^{n-1}}{2^n-1}$$
If $x=1,y=1=y_0$(say)
$\displaystyle\implies\sum_{r=1}^{n-1}y_r=\dfrac{n 2^{n-1}}{2^n-1}-y_0=\dfrac{n 2^{n-1}}{2^n-1}-1=\cdots$