Simplify $\sum^{N-1}_{k=1} \frac{1}{1-\exp(i2\pi k/N)}$

Question

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.

Answer 1

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 $\frac{n-1}{2}$ .

Answer 2

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 $N$.

Answer 3

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 $$2S_N=\sum_{k=1}^{N-1} \left(\frac{1}{1+e^{2ik\pi/N}}+\frac{1}{1+e^{2i(N-k)\pi}}\right)=\sum_{k=1}^{N-1} 1 \implies S_N=\frac{N-1}{2}. $$

Answer 4

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: $\cot(\frac {\frac N2}{N} \pi)=0$

Answer 5

$x_k=e^{2i\pi k/n}, k=0,...n-1$ 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 $$(y-1)^n-(y)^n=0\implies -ny^{n-1}+\frac{n(n-1)}{2}y^{n-2}.....+(-1)^n=0.$$ $y_k,k=1,.....n$ are the roots of the $y-$ equation Hence, the sum $$S_n=\sum_{k=1}^{n-1} \frac{1}{1-e^{2ik\pi/n}}=\sum_{k=1}^{n-1} y_k=-\frac{n(n-1)}{-2n}=\frac{n-1}{2}.$$