Simplify $\sum^{N-1}_{k=1} \frac{1}{1-\exp(i2\pi k/N)}$ I am trying to prove that $$\sum^{N-1}_{k=1} \frac{1}{1-e^{i2\pi k/N}}=\frac{N-1}{2},$$ the closest formula
I can think of is $$\sum^{N-1}_{k=0} r^k = \frac{1-r^N}{1-r},$$
but seems like it is not exactly the case.
 A: Let $U$ be the set of all complex roots of unity $u$ of order $N$, all but $u=1$.
We want to compute the sum over $1/(1-u)$ for $u\in U$.
Consider for this the polynomial $P$:
$$
\begin{aligned}
P(X) &:=\frac{X^N-1}{X-1}=X^{N-1}+\dots+X+1=\prod_{u\in U}(X-u)\ .
\\[3mm]
&\qquad\text{ Then using the product rule}\\
&\qquad\text{  $(fg\dots h)'=f'g\dots h+fg'\dots h+\cdots+fg\dots h$ we get:}\\[3mm]
P'(X)&=\sum_{u\in U}\frac{P(X)}{X-u}\ ,\\
P'(1)&=\sum_{u\in U}\frac{P(1)}{1-u}\ ,\qquad\text{ i.e.}\\
\frac{(N-1)N}2&=\sum_{u\in U}\frac N{1-u}\ ,
\end{aligned}
$$
which is the claimed relation after dividing by $N$.
A: It's a fun problem. The easiest way to solve it is to first note that the set
$$\left\{\mathrm e^{2\pi\mathrm ik/n}\right\}_{k=1,\dots,n-1}$$
Comprise the solutions to the algebraic equation $z^n=1$, other than the trivial solution $z=1$. From our knowledge of complex analysis, we know that these solutions are evenly spread around the unit circle.
Next, we note that the function
$$z\mapsto\frac{1}{1-z}$$
maps the unit circle onto the line $\{z\in\mathbb C\mid \Re(z)=1/2\}$.
To see this let $t\in(0,1)$ and calculate
$$\frac{1}{1-\mathrm e^{2\pi\mathrm it}}=\frac{\overline{1-\mathrm e^{2\pi\mathrm it}}}{|1-\mathrm e^{2\pi\mathrm it}|^2}=\frac{1-\mathrm e^{-\mathrm 2\pi\mathrm it}}{4\sin(\pi t)^2} \\ =\frac{1}{4\sin(\pi t)^2}\big((1-\cos(2\pi t))-\mathrm i\sin(2\pi t)\big) \\ =\frac{1}{4\sin(\pi t)^2}\big(2\sin(\pi t)^2-\mathrm i\cdot 2\sin(\pi t)\cos(\pi t)\big) \\ =\frac{1}{2}+\mathrm i\cdot\frac{1}{2}\cot(\pi t)$$
The key point is that, because the set $\left\{\mathrm e^{2\pi\mathrm ik/n}\right\}_{k=1,\dots,n-1}$ is evenly spread on the unit circle, the set
$$\left\{\frac{1}{1-\mathrm e^{2\pi\mathrm ik/n}}\right\}_{k=1,\dots,n-1}$$
Will be evenly spread on either side of the line $\{z\in\mathbb C\mid \Re(z)=1/2\}$. Therefore, all of the imaginary parts cancel, and our sum is equivalent to summing $n-1$ copies of the number $1/2$. Hence the answer of $\frac{n-1}{2}$
.
A: Let
$$S_N=\sum^{N-1}_{k=1} \frac{1}{1-e^{i2\pi k/N}}=\frac{N-1}{2},$$
Use $$S=\sum_{k=k_1}^{k_2} f(k)=\sum_{k=k_1}^{k_2} f(k_1+k_2-k)\implies 2S= \sum_{k=k_1}^{k_2} [f(k)+f(k_1+k_2-k)]$$
Then we have
$$2S_N=\sum_{k=1}^{N-1} \left(\frac{1}{1+e^{2ik\pi/N}}+\frac{1}{1+e^{2i(N-k)\pi}}\right)=\sum_{k=1}^{N-1} 1 \implies S_N=\frac{N-1}{2}. $$
A: After working out the expression of the sum, it becomes
$$\frac 12\sum_{k=1}^{N-1}{\left [1+i\cot \left(\frac kN \pi \right )\right ]}$$
The reason the imaginary terms cancel out is that since the pair $\cot(\frac kN \pi)=-\cot(\frac{N-k}{N}\pi)$, then

*

*For N odd number, there is an even number of terms terms that pair to cancel each other out

*For N even number, the only term that doesn’t pair to cancel is the middle term, which happens to be zero: $\cot(\frac {\frac N2}{N} \pi)=0$
A: $x_k=e^{2i\pi k/n}, k=0,...n-1$ are $n$ roots of $x^n=1$, Let us transform this equation by $y=\frac{1}{1-x} \implies x=\frac{y-1}{y}$. The transformed equation is $$(y-1)^n-(y)^n\implies -ny^{n-1}+\frac{n(n-1)}{2}y^{n-2}.....+(-1)^n.$$
$y_k,k=1,.....n$ are the roots of the $y-$ equation
Hence, the sum
$$S_n=\sum_{k=1}^{n-1} \frac{1}{1-e^{2ik\pi/n}}=\sum_{k=1}^{n-1} y_k=-\frac{n(n-1)}{-2n}=\frac{n-1}{2}.$$
