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.