TITLE: Mysterious identity
QUESTION [25 upvotes]: Playing around with Maple I found this identity
$$\sum_{k=0}^{n-1}\frac{2k+1}{1-z^{2k+1}}=n\sum_{k=0}^{n-1}\frac{1}{1+z^{k}}$$
where $n$ is a positive integer, $z=\exp(\pi i/n)$. 
I was able to verify it only numerically. Does anyone know how to prove it?

REPLY [3 votes]: Your identity is true. I give here the method for you to prove the point for general $n$. 
One has as main preliminary remark $$\color{red}{z^n=-1\iff z^{n+k}=-z^k\iff \frac{1}{1-z^k}=\frac{z^{n-k}}{1+z^{n-k}}}$$ We make $$A=\sum_{k=o}^{n-1}\frac{2k+1}{1-z^{2k+1}}$$ $$B=n\sum_{k=0}^{n-1}\frac{1}{1+z^k}$$ 
Example of algebraic verification: $n=6$, even.
The odd exponents in $A$ are $\begin{cases}1\\3\\5\\7=6+1\\9=6+3\\11=6+5\end{cases}$
therefore
$$A=\left({1\over 1-z}+{3\over 1-z^3}+{5\over 1-z^5}\right)+\left({6+1\over 1+z}+{6+3\over 1+z^3}+{6+5\over 1+z^5}\right)$$
$$A=B\Rightarrow \left({1\over 1-z}+{3\over 1-z^3}+{5\over 1-z^5}\right)+\left({1\over 1+z}+{3\over 1+z^3}+{5\over 1+z^5}\right)=  6\left({1\over 1+1}+{1\over 1+z^2}+{1\over 1+z^4}\right)$$ Hence $${2\over 1-z^2}+{6\over 1+1}+{10\over 1-z^{10}}=  {6\over 1+1}+{6\over 1+z^2}+{6\over 1+z^4}$$  $${2\over 1-z^2}+{10\over 1+z^4}= {6\over 1+z^2}+{6\over 1+z^4}\iff {1\over 1+z^4}={1-2z^2\over 1-z^4}\iff z^4-z^2+1=0 $$ Since $z^6+1=(z^2+1)(z^4-z^2+1)=0$  the algebraic verification is ended.
Example of algebraic verification: $n=7$, odd.
$$A={1\over 1-z}+{3\over 1-z^3}+{5\over 1-z^5}+{7\over 1+1}+{7+2\over 1+z^2}+{7+4\over 1+z^4}+{7+6\over 1+z^6}=B$$ 
$$\left({1-6z\over 1-z}-{7\over 1+z}+{7z^2\over 1-z^2}+{2+5z^2\over 1+z^2}+{4\over 1+z^4}\right)={7\over 1+z^3}-{3\over 1-z^3}$$
$$\left({2z(2z^7-z^6+z^5-z^4+6z^3-z^2+z-1)\over z^8-1}\right)={7\over 1+z^3}-{3\over 1-z^3}$$ but the parenthesis equals
$${2z(5z^3-2)\over –z-1}$$ because  $$\left({2z(2z^7-(\color{red}{z^6-z^5+z^4-z^3+z^2-z+1})+5z^3)\over z^8-1}\right)={2z(5z^3-2)\over –z-1}$$
where the red polynomial is null as a factor of $z^7+1=0$. Therefore  $${2z(5z^3-2)\over –z-1}={7\over 1+z^3}-{3\over 1-z^3}= {2(5z^2-2)\over z^6-1}$$ i.e $${z\over –z-1}={1\over z^6-1}\iff z^7-z=-z-1$$ which ends the proof $n=7$.