3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ do $n=2$ and $n=3$ explicitly. $\endgroup$ – Will Jagy Feb 10 at 0:33
  • $\begingroup$ @WillJagy, okay, I'm getting that it comes out to $\frac{n-1}{2}$. Thanks. $\endgroup$ – Jake Feb 10 at 0:39
6
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Maybe I'm missing something obvious but could you elaborate on how you determined that the sum of the roots of $R$ is $\frac{n(n-1)/2}{n}$? $\endgroup$ – Alex Feb 10 at 1:15
  • $\begingroup$ 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. $\endgroup$ – zimbra314 Feb 10 at 1:25
3
$\begingroup$

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

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} $

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Jake Feb 10 at 2:34
  • 1
    $\begingroup$ Put n-k for k, reversing the order of summatipn. $\endgroup$ – marty cohen Feb 10 at 3:45
2
$\begingroup$

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.