Let $B\_{n}$ for $n\ge0$ denote the Bernoulli number generated by
\begin{equation\*}
\frac{z}{\textrm{e}^z-1}=\sum\_{n=0}^\infty B\_n\frac{z^n}{n!}=1-\frac{z}2+\sum\_{n=1}^\infty B\_{2n}\frac{z^{2n}}{(2n)!}, \quad \vert z\vert<2\pi.
\end{equation\*}
Could you please help find a reference or give a proof of the identity
\begin{equation}\label{Bernoulli-ID-Qi-csc-csch}\tag{BQID}
\sum\_{j=1}^{2k}(-1)^j\binom{4k+2}{2j}(2^{2j-1}-1)(2^{4k-2j+1}-1)B\_{2j}B\_{4k-2j+2}=0
\end{equation}
for $k\in\mathbb{N}=\{1,2,\dotsc\}$?
Thank you very much.



Ask for a reference or a proof of an identity involving a finite sum and the Bernoulli numbers
Let, $f(x)=\frac{x}{e^x-1}+\frac{x}{2}-1=\frac{x}{2}\coth(\frac{x}{2})-1$


Now the summation in question can be broken in 4 parts. For example, the first part


$$2^{4k}\sum\_{j=1}^{2k}(-1)^j\binom{4k+2}{2j}B\_{2j}B\_{4k-2j+2}$$ is $(4k+2)!$ times the coefficient of $x^{4k+2}$ in $\frac{1}{4}f(2ix)f(2x)$.


Similarly we can compute other terms and get that the total sum is $S=(4k+2)![x^{4k+2}]F(x)$


Where, $F(x)=\frac{1}{4}f(2ix)f(2x)-\frac{1}{2}(f(ix)f(2x)+f(2ix)f(x))+f(x)f(ix)$


Now, $f(x)$ is an even function. So, we can see $F(x)=F(ix)$. But this requires $[x^{4k+2}]F(x)$ to be equal to zero.


