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Question. Is there a simpler way to prove $$B_{2k+1}(1/4) = \frac{-(2k+1) E_{2k}}{4^{2k+1}}$$ where $B_n(x)$ is the $n$-th Bernoulli polynomial and $E_n$ is the $n$-th Euler number?

I have verified it with Mathematica; the identity does not appear anywhere on wikipedia. I have proved it, but the method seems overcomplicated. I used the Fourier series of the Bernoulli polynomials of odd order

$$B_{2r+1}(x) = \frac{(-1)^{r+1} 2 (2r+1)!}{(2\pi)^{2r+1}} \sum_{m \geq 1} \frac{\sin(2\pi m x)}{m^{2r+1}}$$

to derive a closed form expression for odd values of the Dirichlet-beta function. See my answer here. By equating this closed form with the given closed form

$$\beta(2k+1)=\frac{(-1)^kE_{2k}\pi^{2k+1}}{4^{k+1}(2k)!}$$

I obtained the identity at the start. Is there a simpler way of proving this? Thanks

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$(1).$ By the special value of Dirichlet Beta function: $$\beta(-2n-1)=0\quad{\&}\quad\beta(-2n)=\frac{1}{2}E_{2n}$$ $(2).$ By the special value of Hurwitz Zeta function: $$\zeta(-n,a)=-\frac{B_{n+1}(a)}{n+1}\quad{\&}\quad\zeta(-n)=-\frac{B_{n+1}}{n+1}$$ $(3).$ Via Dirichlet Beta representation: $$4^{s}\beta(s)=\zeta{\left(s,\frac14\right)}-\zeta{\left(s,\frac34\right)}$$ $(4).$ Via Hurwitz Zeta multiplication: $$\left(1-2^{-s}\right)4^{s}\zeta(s)=\zeta{\left(s,\frac14\right)}+\zeta{\left(s,\frac34\right)}$$ Hence: $$ \begin{align} \color{red}{B_{n}\left(\frac{1}{4}\right)}&{=-n\,\zeta\left(1-n,\frac14\right)} \\ &{=-n\,\left(4^{(1-n)-{\small1/2}}\right)\left[\left(1-2^{-(1-n)}\right)\,\zeta(1-n)+\beta(1-n)\right]} \\ &{=-n\,\left(4^{{\small1/2}-n}\right)\left[\left(1-2^{n-1}\right)\,\left(-\frac{B_{n}}{n}\right)+\frac{1}{2}E_{n-1}\right]} \\ &\color{red}{=-\frac{n}{4^n}\,E_{n-1}-\frac{1-2^{1-n}}{2^n}\,B_{n}} \end{align} $$ Implies: $$ \begin{align} B_{2k+1}\left(\frac{1}{4}\right)&=-\frac{2k+1}{4^{2k+1}}\,E_{2k} \\ B_{2k}\left(\frac{1}{4}\right)&=-\frac{1-2^{1-2k}}{2^{2k}}\,B_{2k} \end{align} $$

Hazem Orabi
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