Let $n$ be a positive integer and $\zeta$ be a primitive $n$th root of unity. It is not hard to show that
\begin{align\*}
\sum\_{k=1}^{n-1}\frac{\zeta^k}{1-\zeta^k}=\frac{1-n}{2}.
\end{align\*}
Since $\zeta^n=1$, we have
\begin{align\*}
\frac{\zeta^k}{1-\zeta^k}+\frac{\zeta^{n-k}}{1-\zeta^{n-k}}
=\frac{\zeta^k}{1-\zeta^k}+\frac{\zeta^{-k}}{1-\zeta^{-k}}
=-1,
\end{align\*}
and so
\begin{align\*}
\sum\_{k=1}^{n-1}\frac{\zeta^k}{1-\zeta^k}=\frac{1}{2}
\sum\_{k=1}^{n-1}\left(\frac{\zeta^k}{1-\zeta^k}+\frac{\zeta^{-k}}{1-\zeta^{-k}}\right)
=\frac{1-n}{2}.
\end{align\*}
Let $\omega$ be a primitive $(3n+2)$th root of unity. By the same method, we can also get
\begin{align\*}
\sum\_{k=0}^{2n+1}(-1)^k\omega^{\frac{k(3k+1)}{2}}=
\sum\_{k=0}^{n}\left((-1)^k\omega^{\frac{k(3k+1)}{2}}+(-1)^{2n+1-k}
\omega^{\frac{(2n+1-k)(3(2n+1-k)+1)}{2}}\right)=0,
\end{align\*}
because
\begin{align\*}
\omega^{\frac{(2n+1-k)(3(2n+1-k)+1)}{2}}=\omega^{\frac{k(3k+1)}{2}+(3n+2)(2n-2k+1)}=
\omega^{\frac{k(3k+1)}{2}}.
\end{align\*}
Note that the above two sums possess natural symmetries.


**Question.**
Let $\omega=e^\frac{2\pi i}{3n+2}$ be the primitive $(3n+2)$th root of unity.
Numerical calculation suggests that
\begin{align\*}
\sum\_{k=1}^{2n+1}\frac{(-1)^k\omega^{\frac{k(3k+1)}{2}}}{1-\omega^{3k}}=-\frac{n+1}{2}.
\end{align\*}
Is this identity true? If so, how to prove it?
Unfortunately, this sum loses a natural symmetry.
Hints, references or proofs are all welcome!


**Comments:**
Nemo proved that for arbitrary $w$,
\begin{align\*}
\sum \_{k=1}^{2 n+1} \frac{(-1)^k w^{k (3 k+1)/2}}{1-w^{3 k}}&=-\sum \_{k=0}^{2 n} \frac{(-1)^k w^{\frac{1}{2} (k+2) (3 k+1)}}{1-w^{3 k+1}}\\
&+\frac{1}{2} \sum \_{k=0}^{2 n} \left(\frac{(-1)^k w^{(3 k+1) (n+1)}}{1-w^{(3k+1)/2}}+\frac{w^{(3 k+1) (n+1)}}{w^{(3k+1)/2}+1}\right)\\
&+\frac{1}{2} \sum \_{k=1}^{2 n+1} \left(\frac{(-1)^k w^{k/2}}{1-w^{3k/2}}+\frac{w^{k/2}}{w^{3k/2}+1}\right).
\end{align\*}
This identity has been reproved by GH from MO below.


Letting $k\to 2n+1-k$ in the following sum gives
\begin{align\*}
\sum \_{k=0}^{2 n} \frac{(-1)^k w^{\frac{1}{2} (k+2) (3 k+1)}}{1-w^{3 k+1}}
=\sum \_{k=1}^{2 n+1} \frac{(-1)^k w^{k (3 k+1)/2}}{1-w^{3 k}},
\end{align\*}
where we have used the fact $w^{3n+2}=1$. Also, we have (letting $k\to 2n+1-k$)
\begin{align\*}
 \sum \_{k=0}^{2 n} \left(\frac{(-1)^k w^{(3 k+1) (n+1)}}{1-w^{(3k+1)/2}}+\frac{w^{(3 k+1) (n+1)}}{w^{(3k+1)/2}+1}\right)=
 \sum \_{k=1}^{2 n+1} \left(\frac{(-1)^k w^{k/2}}{1-w^{3k/2}}+\frac{w^{k/2}}{w^{3k/2}+1}\right).
\end{align\*}
So it suffices to show that 
\begin{align\*}
\sum \_{k=1}^{2 n+1} \left(\frac{(-1)^k w^{k/2}}{1-w^{3k/2}}+\frac{w^{k/2}}{w^{3k/2}+1}\right)=-n-1,
\end{align\*}
which was proved by Fedor Petrov.



A sum involving roots of unity
Here is the proof of Kevin Liu's version 
$$
\sum\_{k=1}^{2n+1}\left(\frac{(-y)^k}{1-y^{3k}}+ \frac{y^{k}}{1+y^{3k}}\right)=-n-1
$$
(for the primitive root of unity $y$ of degree $6n+4$) of Nemo's reduction. (Both reductions deserve to be explained, in my opinion.) 


We have $$\sum\_{k=1}^{2n+1} \frac{(-y)^k}{1-y^{3k}}=
-\sum\_{k=1}^{2n+1} \frac{y^{k}}{1-y^{3k}}+2\sum\_{k=1}^n\frac{y^{2k}}{1-y^{6k}}.$$


So we should prove


$$2\sum\_{k=1}^{2n+1} \frac{y^{4k}}{1-y^{6k}}-2\sum\_{k=1}^n \frac{y^{2k}}{1-y^{6k}}=n+1$$


Denote again $w=y^2$, the primitive root of unity of degree $3n+2$, this reads as


$$
2\sum\_{k=1}^{2n+1}\frac{w^{2k}}{1-w^{3k}}-2\sum\_{k=1}^n\frac{w^k}{1-w^{3k}}=n+1
$$


Partition LHS onto two identical halfs (it has multiple 2 for that), and in one of them make the change of variables $k\mapsto 3n+2-k$. This half reads as 
$$
\sum\_{k=n+1}^{3n+1}\frac{w^{-2k}}{1-w^{-3k}}-\sum\_{k={2n+2}}^{3n+1}\frac{w^{-k}}{1-w^{-3k}}=
-\sum\_{k=n+1}^{3n+1}\frac{w^{k}}{1-w^{3k}}+\sum\_{k={2n+2}}^{3n+1}\frac{w^{2k}}{1-w^{3k}}.
$$
Collecting with another half we get (so lucky)
$$
\sum\_{k=1}^{3n+1} \frac{w^{2k}-w^k}{1-w^{3k}}=-\sum\_{k=1}^{3n+1} w^k\frac{1-w^{3k(n+1)}}{1-w^{3k}}=
-\sum\_{k=1}^{3n+1} (w^k+w^{4k}+w^{7k}+\dots+w^{(3n+1)k})=n+1,
$$
since the numbers $1,4,\dots,3n+1$ are not divisible by $3n+2$ and $\sum\_{k=0}^{3n+1} w^{kd}=0$ for all integers $d$ non-divisible by $3n+2$.


