TITLE: A sum involving roots of unity
QUESTION [27 upvotes]: 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.

REPLY [4 votes]: Here is a proof of Nemo's identity. Using the notation $y:=w^{1/2}$, multiplying both sides by $2$, and shifting $k$ by $1$ in three of the five sums, the identity can be rewritten as
$$\sum_{k=1}^{2n+1}f_{k,n}(y)=0,$$
where
\begin{align*}f_{k,n}(y):=&\frac{(-1)^ky^k}{1-y^{3k}}+\frac{y^k}{1+y^{3k}}-2\frac{(-1)^ky^{k(3k+1)}}{1-y^{6k}}\\[6pt]
&+2\frac{(-1)^ky^{(k+1)(3k-2)}}{1-y^{6k-4}}-\frac{(-1)^ky^{(2n+2)(3k-2)}}{1-y^{3k-2}}
+\frac{y^{(2n+2)(3k-2)}}{1+y^{3k-2}}\\[6pt]
=&(-1)^ky^k\frac{1-y^{3k^2}}{1-y^{3k}}+y^k\frac{1-(-1)^ky^{3k^2}}{1+y^{3k}}\\[6pt]
&+(-1)^ky^{(k+1)(3k-2)}\frac{1-y^{(2n-k+1)(3k-2)}}{1-y^{3k-2}}
+(-1)^ky^{(k+1)(3k-2)}\frac{1+(-1)^ky^{(2n-k+1)(3k-2)}}{1+y^{3k-2}}\\[6pt]
=&\sum_{m=0}^{k-1}\left((-1)^k+(-1)^m\right)y^{k(3m+1)}+
\sum_{m=0}^{2n-k}\left((-1)^k+(-1)^{k+m}\right)y^{(k+m+1)(3k-2)}.
\end{align*}
In the first $m$-sum, the term $m=k-1$ does not contribute, hence what we really need to prove is
$$\sum_{k=1}^{2n+1}\sum_{m=0}^{k-2}\left((-1)^k+(-1)^m\right)y^{k(3m+1)}+
\sum_{k=1}^{2n+1}\sum_{m=0}^{2n-k}\left((-1)^k+(-1)^{k+m}\right)y^{(k+m+1)(3k-2)}
=0.$$
In the second double sum, we make the change of variables $k':=k+m+1$ and $m':=k-1$. With this notation, the previous equation becomes
$$\sum_{k=1}^{2n+1}\sum_{m=0}^{k-2}\left((-1)^k+(-1)^m\right)y^{k(3m+1)}+
\sum_{k'=1}^{2n+1}\sum_{m'=0}^{k'-2}\left((-1)^{m'+1}+(-1)^{k'-1}\right)y^{k'(3m'+1)}
=0.$$
The two double sums now clearly neutralize each other termwise, and the proof is complete.