Compute $\sum\frac1{2-A_k}$ for $(A_k)$ the $n$th roots of unity

Question

If $1,A_1,A_2,A_3....A_{n-1}$ are the $n^{th}$ roots of unity then prove that $$\dfrac{1}{2-A_1} + \dfrac{1}{2-A_2}+\cdots+ \dfrac{1}{2-A_{n-1}} = \dfrac{2^{n-1}(n-2) + 1}{2^n-1}$$

What I did: I tried to use some of the following formulas: $$1+ A_1 +A_2+A_3+\cdots+A_{n-1} = 0$$

$$\dfrac{2^n - 1}{2-1} = (2 -A_1)(2-A_2)\cdots(2-A_{n-1})$$

and the fact that $|A_i| = 1$ for $i =1,2,3,\cdots,n-1$.

Answer 1

Note that $$ z^n-1=\prod_{k=0}^{n-1}(z-a_k) $$ so that $$ nz^{n-1}=\sum_{k=0}^{n-1}\frac1{z-a_k}\prod_{k=0}^{n-1}(z-a_k) $$ and therefore, $$ \frac{nz^{n-1}}{z^n-1}=\sum_{k=0}^{n-1}\frac1{z-a_k} $$ Thus, $$ \sum_{k=0}^{n-1}\frac1{2-a_k}=\frac{n2^{n-1}}{2^n-1} $$ Subtracting the $k=0$ ($a_0=1$) term gives $$ \sum_{k=1}^{n-1}\frac1{2-a_k}=\frac{(n-2)2^{n-1}+1}{2^n-1} $$

Answer 2

Two ingredients for this proof:

• The numbers $A_k$ for $1\leqslant k<n$ and $A_n=1$ are the roots of the polynomial $X^n-1$.
• If $Y=\frac1{2-X}$ then $X=\frac{2Y-1}Y$.

Thus, the numbers $B_k=\frac1{2-A_k}$ for $1\leqslant k<n$ and $B_n=\frac1{2-A_n}=1$ are the solutions of the equation $$\left(\frac{2Y-1}Y\right)^n-1=0$$ that is, the roots of the polynomial $$(2Y-1)^n-Y^n$$

The highest degree terms of this polynomial are $$(2^n-1)Y^n-n2^{n-1}Y^{n-1}+\ldots$$ hence the sum of its roots is $$\sum_{k=1}^nB_k=1+\sum_{k=1}^{n-1}\frac1{2-A_k}=\frac{n2^{n-1}}{2^n-1}$$ from which the desired result follows.