$\sum_{k=1}^{n-1}\left(1-e^{\frac{2\pi ki}{n}}\right)^{-1}$

Question

How can I go about computing $$ \sum_{k\ =\ 1}^{n - 1} \left(1 - \mathrm{e}^{\large 2\pi k\mathrm{i}/n}\right)^{-1}\ {\Large ?} $$

I originally thought that it was supposed to be the reciprocal of the sum, and I ended up with $1/n$, but now I realized that it is the sum of the reciprocals. I've tried using $$e^{ix}=\cos x+i\sin x,$$ but I didn't get anywhere with that.

Answer 1

The answer is $\frac{n-1}{2}$. There are multiple ways to approach it, here is one way;

Let $\displaystyle \alpha_k = e^{i \frac{2\pi k}{n}}$ for $k = 1, \dots, n-1$. We seek to evaluate,

$$\sum_{k=1}^{n-1} \frac{1}{1-\alpha_k}, \ \ (*) $$

We know that $\alpha_1, \dots , \alpha_{n-1}$ are roots of the polynomial,

$$P(z) = 1 + z+ \dots + z^{n-1}$$

This is because $1, \alpha, \dots, \alpha_{n-1}$ are the roots of the polynomial $1 - z^n$ (roots of unity), so when we take away $1$ as a root we obtain $P$. But $\alpha_k$ is a root of $P$ simply means that $P(\alpha_k) = 0$ for each $k$. Our goal now is to find a polynomial that has roots,

$$\frac{1}{1-\alpha_k} = f(\alpha_k) $$

as roots instead of $\alpha_k$, with this polynomial, we find the sum of the roots of that polynomial to obtain $(*)$. Since $f$ is one-to-one we can find the inverse function,

$$f^{-1}(x) = 1 - \frac{1}{x} \Rightarrow f^{-1} \left(\frac{1}{1-\alpha_k} \right) = \alpha_k$$

Therefore the function,

$$Q(x) = P(f^{-1}(x)) = P \left(1 - \frac{1}{x}\right) $$

has roots $\displaystyle \frac{1}{1-\alpha_k}$ for each $k$ (this is the crux of the argument, please convince yourself of this!). Note that $Q$ as defined is not a polynomial yet, but we can use $Q$ as a function to find a polynomial that has the same roots.

Now, we have that,

$$Q(x) = P \left(1 - \frac{1}{x} \right) = 1 + \left(1 - \frac{1}{x} \right) + \dots + \left( 1- \frac{1}{x} \right)^{n-1} $$

Summing this geometric series we obtain,

$$Q(x) = x\left(1 - \left(1-\frac{1}{x} \right)^n\right) $$

With some manipulation we obtain that,

$$R(x) = x^{n-1}Q(x) = x^n - (x-1)^n = nx^{n-1} - \frac{n(n-1)}{2} x^{n-2} + \dots $$

Now we have a polynomial $R$ and I claim that $R$ has roots $\displaystyle \frac{1}{1-\alpha_k}$ for each $k$. This is because,

$$R \left(\frac{1}{1-\alpha_k} \right) = \frac{1}{(1-\alpha_k)^{n-1}} Q \left(\frac{1}{1-\alpha_k} \right) =\frac{1}{(1-\alpha_k)^{n-1}} P\left(f^{-1}\left(\frac{1}{1-\alpha_k} \right) \right) = \frac{1}{(1-\alpha_k)^{n-1}} P(\alpha_k) = 0 $$

Therefore to find $(*)$ we note that it is simply the sum of the roots of $R$, therefore,

$$ \sum_{k=1}^{n-1} \frac{1}{1-\alpha_k} = \frac{n(n-1)/2}{n} = \frac{n-1}{2}$$

This completes the proof.

Answer 2

$\sum_{k=1}^{n-1}\left(1-e^{\frac{2\pi ki}{n}}\right)^{-1}$

I will show that the sum is $\dfrac{n-1}{2} $.

This is undoubtedly well-known, but it wasn't to me until I did this.

Using your suggestion, since $\frac1{a+bi} =\frac{a-bi}{a^2+b^2} $,

$\begin{array}\\ \left(1-e^{\frac{2\pi ki}{n}}\right)^{-1} &=\left(1-\cos(2\pi k /n)-i\sin(2\pi k/n)\right)^{-1}\\ &=\dfrac{1-\cos(2\pi k /n)+i\sin(2\pi k/n)}{(1-\cos(2\pi k /n))^2+\sin^2(2\pi k/n)}\\ &=\dfrac{1-\cos(2\pi k /n)+i\sin(2\pi k/n)}{1-2\cos(2\pi k /n)+\cos(2\pi k /n)^2+\sin^2(2\pi k/n)}\\ &=\dfrac{1-\cos(2\pi k /n)+i\sin(2\pi k/n)}{2-2\cos(2\pi k /n)}\\ &=\dfrac{1-\cos(2\pi k /n)+i\sin(2\pi k/n)}{2(1-\cos(2\pi k /n))}\\ &=\dfrac12+i\dfrac{\sin(2\pi k/n)}{2(1-\cos(2\pi k /n))}\\ &=\dfrac12+i\dfrac{2\sin(\pi k/n)\cos(\pi k/n)}{2(2\sin^2(\pi k /n)))}\\ &=\dfrac12+\dfrac{i}{2}\dfrac{\cos(\pi k/n)}{\sin(\pi k /n)}\\ &=\dfrac12+\dfrac{i}{2}\cot(\pi k/n)\\ \end{array} $

so that

$\begin{array}\\ \sum_{k=1}^{n-1}\left(1-e^{\frac{2\pi ki}{n}}\right)^{-1} &=\sum_{k=1}^{n-1}\left(\dfrac12+\dfrac{i}{2}\cot(\pi k/n) \right)\\ &=\dfrac{n-1}{2}+\dfrac{i}{2}\sum_{k=1}^{n-1}\cot(\pi k/n) \\ \end{array} $

However, if $S =\sum_{k=1}^{n-1}\cot(\pi k/n) $, then $S =\sum_{k=1}^{n-1}\cot(\pi (n-k)/n) =\sum_{k=1}^{n-1}\cot(\pi-\pi k/n) $, so that $2S =\sum_{k=1}^{n-1}(\cot(\pi k/n)+\cot(\pi-\pi k/n)) =0 $ since

$\begin{array}\\ \cot(x)+\cot(\pi-x) &=\dfrac{\cos(x)}{\sin(x)}+\dfrac{\cos(\pi-x)}{\sin(\pi-x)}\\ &=\dfrac{\cos(x)}{\sin(x)}+\dfrac{-\cos(x)}{\sin(x)}\\ &=0\\ \end{array} $

Answer 3

With $\zeta = \exp(2\pi i/n)$ and

$$f(z) = \frac{1}{1-z} \frac{n/z}{z^n-1}$$

we have for $1\le k\le n-1$

$$\mathrm{Res}_{z=\zeta^k} f(z) = \frac{1}{1-\zeta^k}$$

so that

$$S = \sum_{k=1}^{n-1} \frac{1}{1-\zeta^k} = \sum_{k=1}^{n-1} \mathrm{Res}_{z=\zeta^k} f(z).$$

Residues sum to zero and the residue at infinity is zero, so we find

$$S = - \mathrm{Res}_{z=1} f(z) - \mathrm{Res}_{z=0} f(z).$$

For the first one we have

$$- \mathrm{Res}_{z=1} f(z) = \mathrm{Res}_{z=1} \frac{1}{z-1} \frac{n/z}{z^n-1} \\ = \mathrm{Res}_{z=1} \frac{1}{(z-1)^2} \frac{n/z}{1+z+\cdots+z^{n-1}} \\ = n \left.\left(\frac{1}{z} \frac{1}{1+z+\cdots+z^{n-1}} \right)'\right|_{z=1} \\ = n \left.\left(- \frac{1}{z^2} \frac{1}{1+z+\cdots+z^{n-1}} - \frac{1}{z} \frac{(1+\cdots+(n-1) z^{n-2})} {(1+z+\cdots+z^{n-1})^2} \right)\right|_{z=1} \\ = n \left( - \frac{1}{n} - \frac{1}{n^2} \frac{1}{2} (n-1) n \right) = -1 - \frac{1}{2} (n-1).$$

The second one is

$$- \mathrm{Res}_{z=0} f(z) = - (1 \times n \times -1) = n.$$

Collecting everything we get

$$\bbox[5px,border:2px solid #00A000]{ S = \frac{1}{2} (n-1).}$$

Answer 4

Call $\omega = \exp(2\pi\text i/n)$. You have $$ \sum_{k=1}^{n-1} \frac 1{1-\omega^k} = \sum_{k=1}^{n-1} \frac 1{1-\omega^{n-k}}\\ \implies \sum_{k=1}^{n-1} \frac 1{1-\omega^k} = \frac 12 \sum_{k=1}^{n-1} \frac 1{1-\omega^k} + \frac 1{1-\omega^{n-k}}\\ = \frac 12 \sum_{k=1}^{n-1} \frac {2-\omega^k-\omega^{n-k}}{2-\omega^k-\omega^{n-k}} =\frac{n-1}2 $$