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

How can I go about computing $$\sum_{k=1}^{n-1}\left(1-e^{\frac{2\pi ki}{n}}\right)^{-1}?$$

I originally thought that it was supposed to be the reciprocal of the sum, and I ended up with $$\frac{1}{n}$$, 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.

• do $n=2$ and $n=3$ explicitly. – Will Jagy Feb 10 at 0:33
• @WillJagy, okay, I'm getting that it comes out to $\frac{n-1}{2}$. Thanks. – Jake Feb 10 at 0:39

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.

• Maybe I'm missing something obvious but could you elaborate on how you determined that the sum of the roots of $R$ is $\frac{n(n-1)/2}{n}$? – Alex Feb 10 at 1:15
• Great answer. @Alex, sum of roots of a polynomial $a_{k}x^{k}+a_{k-1}x^{k-1}+\cdots a_{0}=0$ is $-a_{k-1}/a_{k}$ also known as Vieta's formulas. – zimbra314 Feb 10 at 1:25

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

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

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

• I'm following everything up until $\sum_{k=1}^{n-1}\cot(\pi k/n)=\sum_{k=1}^{n-1}\cot(\pi (n-k)/n)$. How are you getting to that? – Jake Feb 10 at 2:34
• Put n-k for k, reversing the order of summatipn. – marty cohen Feb 10 at 3:45

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

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