Ask for a reference or a proof of an identity involving a finite sum and the Bernoulli numbers 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.
 A: 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.
A: On page 42 in the handbook [1] listed as a reference below, the series expansions
\begin{equation}\label{csc-ser-eq}
\csc x=\frac1x+\sum_{k=1}^\infty\frac{2\bigl(2^{2k-1}-1\bigr)|B_{2k}|}{(2k)!}x^{2k-1}
\end{equation}
and
\begin{equation}\label{csch-ser-eq}
\textrm{csch}\,x=\frac1x-\sum_{k=1}^\infty\frac{2\bigl(2^{2k-1}-1\bigr)B_{2k}}{(2k)!}x^{2k-1}
\end{equation}
for $x\in(-\pi,\pi)$ are collected. Hence, by the Cauchy product of two infinite series in mathematical analysis, we obtain
\begin{align*}
x^2\csc x\textrm{csch}\,x&=(x\csc x)(x\textrm{csch}\,x)\\
&=\Biggl[1+\sum_{k=1}^\infty\frac{2\bigl(2^{2k-1}-1\bigr)|B_{2k}|}{(2k)!}x^{2k}\Biggr]
\Biggl[1-\sum_{k=1}^\infty\frac{2\bigl(2^{2k-1}-1\bigr)B_{2k}}{(2k)!}x^{2k}\Biggr]\\
&=1+2\sum_{k=2}^{\infty}\Biggl[\bigl(2^{2k-1}-1\bigr)(|B_{2k}|-B_{2k})\\
&\quad-2\sum_{j=1}^{k-1}\binom{2k}{2j}\bigl(2^{2j-1}-1\bigr) \bigl(2^{2k-2j-1}-1\bigr)|B_{2j}|B_{2k-2j}\Biggr] \frac{x^{2k}}{(2k)!}
\end{align*}
for $x\in(-\pi,\pi)$.
Since the function $x\csc\sqrt{x}\,\textrm{csch}\,\sqrt{x}\,$ is even on the interval $\bigl(-\pi^2,\pi^2\bigr)$, or say, the function $f(x)=x^2\csc x\textrm{csch}\,x$ satisfies $f(x)=f(x\textrm{i})$ on $(-\pi,\pi)$, the identity \eqref{Bernoulli-ID-Qi-csc-csch} follows readily.
This proof is excerpted from the first proof of Theorem 2.1 in the paper [2] below.
Reference

*

*I. S. Gradshteyn and I. M. Ryzhik, Table of Integrals, Series, and Products, Translated from the Russian, Translation edited and with a preface by Daniel Zwillinger and Victor Moll, Eighth edition, Revised from the seventh edition, Elsevier/Academic Press, Amsterdam, 2015; available online at https://doi.org/10.1016/B978-0-12-384933-5.00013-8.

*Xue-Yan Chen, Lan Wu, Dongkyu Lim, and Feng Qi, Two identities and closed-form formulas for the Bernoulli numbers in terms of central factorial numbers of the second kind, Demonstratio Mathematica 55 (2022), no. 1, 822--830; available online at https://doi.org/10.1515/dema-2022-0166.

