Mysterious identity 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?

Given positive integer $n$ that defines $z := \exp(\pi i/n),$ we can generalize the oddly periodic sum with
$$
S_n(x) := \sum_{k=0}^{n-1}\frac{2k+1}{1-xz^{2k+1}}.
$$
The power series when $|x| < 1$ is
$$
S_n(x) = \sum_{E=0}^{\infty}x^E\sum_{k=0}^{n-1}(2k+1)z^{2kE+E}.
$$
This simplifies with the weighted sum,
$$
\left.\frac{d}{dr}(\frac{r-r^{2n+1}}{1-r^2})\right|_{r=z^E} = 
\sum_{k=0}^{n-1}(2k+1)z^{2kE},\  \mbox{ to}
$$ $$
S_n(x) = \sum_{E=0}^{\infty}\frac{1-z^{2E}-(2n+1)z^{2En}+(2n+1)z^{2En+2E}+
2z^{2E}-2z^{2En+2E}}{(1-z^{2E})^2}(xz)^E. 
$$
We observe that $z^{2E}\to1$ correctly reduces the fraction to $n^2,\  $ the period is (almost) $n,\  $ and
$$
S_n(x) = \frac{1}{1+x^n}\left( n^2 - 2n\sum_{E=1}^{n-1}
\frac{(xz)^E}{1-z^{2E}} \right).
$$
Before we take $x\to1,\  $ we need to recall
$$
z=\exp(\pi i/n)\  \implies\  n = 1+\sum_{E=1}^{n-1}\frac{2}{1-z^{2E}}.
$$ 
We can evaluate our analytic sum at
$$
S_n(1) = \frac{n}{2}\left(1 + 2\sum_{E=1}^{n-1}\frac{1-z^E}{1-z^{2E}}
\right)
$$
to find the desired relationship
$$
\sum_{k=0}^{n-1}\frac{2k+1}{1-z^{2k+1}}\  =\  
\sum_{k=0}^{n-1}\frac{n}{1+z^k}.
$$
