# interesting series involving sin and nth roots of unity.

I ran across an interesting series involving the square of sine. It was solved in a clever manner I do not quite get.

$$\sum_{k=0}^{n-1}\frac{1}{1+8\sin^{2}(\frac{k\pi}{n})}$$

The solution involved the nth roots of unity.

$$\sin^{2}\left(\frac{k\pi}{n}\right)=\frac{2-L^{k}-L^{-k}}{4},$$ where $$L=\displaystyle e^{\frac{2\pi i}{n}}$$

Which becomes $$\frac{1}{1+8\sin^{2}(\frac{k\pi}{n})}=\frac{1}{3}\left[\frac{1}{2L^{k}-1}-\frac{2}{L^{k}-2}\right]$$

Noting that $$L^{k}, \;\ (0\leq k\leq n-1)$$ are the nth roots of unity, the following series are used:

$$\sum_{k=0}^{n-1}\frac{1}{2L^{k}-1}=\frac{n}{2^{n}-1}$$ and

$$\sum_{k=0}^{n-1}\frac{1}{L^{k}-2}=\frac{n2^{n-1}}{1-2^{n}}.$$

My question is, how are those last two closed form for the series evaluated?

This much is explained: By noting that $$z^{n}=1$$ and using $$y=\frac{1}{2z-1}, \;\ z=\frac{y+1}{2y},$$

we can construct $$\left(\frac{y+1}{2y}\right)^{n}=1.$$

Then, $$(y+1)^{n}=2^{n}y^{n}.\tag1$$ Then, $$(1-2^{n})y^{n}+ny^{n-1}+\cdot\cdot\cdot +1=0.\tag2$$

So, from this last equation we get $$\sum_{k=0}^{n-1}y_{k}=\sum_{k=0}^{n-1}\frac{1}{2L^{k}-1}=\frac{n}{2^{n}-1},$$ where $$y_{k}$$ are the roots of the equation in $(2).$

My problem is, I still do not quite understand that last line and how that closed form is found. I get down to $(1)$ and after that get lost. Is that a binomial series, and how is it derived from the previous step?.

and how does $$\sum_{k=0}^{n-1}y_{k}=\sum_{k=0}^{n-1}\frac{1}{2L^{k}-1}=\frac{n}{2^{n}-1}$$

Apparently, $$y_{k}=\frac{1}{2L^{k}-1}$$ somehow. Thanks very much for any input. I always like to learn something new and there is probably something here to learn.

I suppose there is something about nth roots of unity I do not get as of yet.

-

The explanation is written in a recondite manner, but its essence is not so hard to grasp. Note that

$$y_k = \frac{1}{2L^k - 1}$$

form $n$ solutions of the equation

$$\left( \frac{y+1}{2y} \right)^n = 1 \quad \Longleftrightarrow \quad 2^n y^n - (y+1)^n = 0.$$

Since $L^k \neq L^l$ if $k$ and $l$ are different mod $n$, we find that $y_0, \cdots, y_{n-1}$ are all different. But the last equation if of degree $n$, so it has at most $n$ zeros. That is, we must have

$$2^n y^n - (y+1)^n = (2^n - 1)(y - y_0) \cdots (y - y_{n-1}),$$

where the proceeding term $2^n - 1$ in the RHS arises since the leading coefficient of the LHS is $2^n - 1$. Expanding the LHS by binomial theorem, we have

$$2^n y^n - (y+1)^n = 2^n y^n - (y^n + ny^{n-1} + \cdots) = (2^n - 1)y^n - n y^{n-1} + \cdots ,$$

and expanding the RHS yields

$$(2^n - 1)(y - y_0) \cdots (y - y_{n-1}) = (2^n - 1)y^n - (2^n - 1)(y_0 + \cdots + y_{n-1})y^{n-1} + \cdots.$$

Therefore we obtain the desired relation

$$y_0 + \cdots + y_{n-1} = \frac{n}{2^n - 1}.$$

This is quite ingenious and simple, but you can also obtain this formula more naturally as follows: Since

$$\sum_{k=0}^{n-1} \frac{1}{2L^k - 1} = \sum_{k=0}^{n-1} \frac{\frac{L^{-k}}{2}}{1 - \frac{L^{-k}}{2}} = \sum_{k=0}^{n-1} \sum_{l=1}^{\infty} \left( \frac{L^{-k}}{2} \right)^{l} = \sum_{l=1}^{\infty} \sum_{k=0}^{n-1} \left( \frac{L^{-k}}{2} \right)^{l}$$

and

$$\sum_{k=0}^{n-1} L^{km} = \begin{cases} n & m \equiv 0 \ (\mathrm{mod} \ n) \\ 0 & \mathrm{otherwise} \end{cases},$$

we must have

$$\sum_{k=0}^{n-1} \frac{1}{2L^k - 1} = \sum_{l'=1}^{\infty} \frac{n}{2^{nl'}} = \frac{n}{2^n - 1},$$

where $l = nl'$. Similar consideration also works for the second one.

-
Thanks a lot sos. I appreciate the explanation. I can see there were details left out. –  Cody Mar 31 '12 at 18:22