Closed form for the sums $S(n)={(2n-1)!\over \sqrt2}\cdot{\left(4/ \pi\right)^{2n}}\cdot\sum\limits_{k=0}^{\infty}(-1)^{k(k+1)/2}(2k+1)^{-2n}$ Consider the sums

$$S(n)={(2n-1)!\over \sqrt2}\cdot{\left(4\over \pi\right)^{2n}}\cdot\sum_{k=0}^{\infty}{(-1)^{k(k+1)\over 2}\over (2k+1)^{2n}}$$

We have $S(1)=1$, $S(2)=11$, $S(3)=361$, $S(4)=24611$.
I cannot spot any pattern within this sequence.
How can we work out a closed form for $S(n)$?
 A: We may notice that
$$ T(n)=\sum_{k\geq 0}\frac{(-1)^{k(k+1)/2}}{(2k+1)^{2n}} = \sum_{m\geq 1}\frac{\chi(m)}{m^{2n}} = L(\chi,2n)\tag{1}$$
is a Dirichlet $L$-function associated with the multiplicative function $\chi(m)$, that equals $0$ if $m$ is even, $1$ if $m\equiv \pm 1\pmod{8}$ and $-1$ if $m\equiv \pm 3\pmod{8}$. In particular
$$ T(n) = \prod_{p>2}\left(1-\frac{\left(\frac{2}{p}\right)}{p^{2n}}\right)^{-1}\tag{2} $$
where $\left(\frac{2}{p}\right)$ is Legendre's symbol. From Hazem Orabi's integral representation
$$ S(n) = \frac{1}{\sqrt{2}(2\pi)^{2n}}\int_{0}^{+\infty}\frac{x^{2n-1}}{e^x-1}\left(e^{x/8}-e^{3x/8}-e^{5x/8}+e^{7x/8}\right)\,dx \tag{3} $$
we also have:
$$ S(n) = \frac{1}{\sqrt{2}}\left(\frac{4}{\pi}\right)^{2n}\int_{1}^{+\infty}\log(x)^{2n-1}\frac{x^2-1}{1+x^4}\,dx \tag{4}$$
or
$$ S(n) = \frac{1}{\sqrt{2}}\left(\frac{4}{\pi}\right)^{2n}\left.\frac{d^{2n-1}}{d\alpha^{2n-1}}\int_{1}^{+\infty}\frac{x^{2+\alpha}-x^{\alpha}}{1+x^4}\,dx\, \right|_{\alpha=0^+}\tag{5}$$
so $S(n)$ depends on $\psi^{(2n-1)}(z)$ (the $(2n-1)$-th derivative of the digamma function, i.e. the $2n$-th derivative of $\log\Gamma$) evaluated at $z=\frac{1}{8},\frac{3}{8},\frac{5}{8},\frac{7}{8}$. By the reflection formulas for the $\psi^{(2n-1)}(z)$ function, everything boils down to the derivatives of the function $\cot(\pi z)$ at $z=\frac{1}{8}$ and $z=\frac{3}{8}$. This proves G.H.Hardy's claim: 

$$ S(n) = (2n-1)!\cdot [z^{2n-1}]\frac{\sin(z)}{\cos(2z)}.\tag{6} $$

In particular, $S(n)$ can be expressed in terms of Euler's polynomials $E_n(z)$ evaluated at $z=\frac{1}{4}$ and $z=\frac{3}{4}$.
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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\begin{align}
\mrm{S}\pars{n} & \equiv
{\pars{2n - 1}! \over \root{2}}\,\pars{4 \over \pi}^{2n}\sum_{k = 0}^{\infty}
{\pars{-1}^{k\pars{k + 1}/2} \over \pars{2k + 1}^{2n}}
\\[5mm] & =
\pars{2n - 1}!\,{2^{4n - 1/2} \over \pi^{2n}}\bracks{%
\sum_{k = 0}^{\infty}{\pars{-1}^{k} \over \pars{4k + 1}^{2n}} +
\sum_{k = 0}^{\infty}{\pars{-1}^{k + 1} \over \pars{4k + 3}^{2n}}}
\\[5mm] & =
{\root{2} \over 2\pi^{2n}}\bracks{%
\pars{2n - 1}!\sum_{k = 0}^{\infty}{\pars{-1}^{k} \over \pars{k + 1/4}^{2n}} -
\pars{2n - 1}!\,\sum_{k = 0}^{\infty}{\pars{-1}^{k} \over \pars{k + 3/4}^{2n}}}
\label{1}\tag{1}
\end{align}

However,

\begin{align}
&\left.\pars{2n - 1}!\,\sum_{k = 0}^{\infty}{\pars{-1}^{k} \over \pars{k + a}^{2n}}\right\vert_{\ a\ >\ 0} =
\pars{2n - 1}!\,\sum_{k = 0}^{\infty}\pars{-1}^{k}\bracks{%
{1 \over \Gamma\pars{2n}}
\int_{0}^{\infty}t^{2n - 1}\expo{-\pars{k + a}t}\,\dd t}
\\[5mm] = &\
\int_{0}^{\infty}t^{2n - 1}\expo{-at}\sum_{k = 0}^{\infty}
\pars{-\expo{-t}}^{k}\,\dd t =
\int_{0}^{\infty}t^{2n - 1}\expo{-at}\,{1 \over 1 + \expo{-t}}\,\dd t
\label{2}\tag{2}
\\[5mm] = &\
\bbx{\ds{4^{-n}\,\Gamma\pars{2n}\bracks{%
\zeta\pars{2n,{a \over 2}} - \zeta\pars{2n,{a + 1 \over 2}}}}}
\end{align}

The last integral, in \eqref{2}, is straightforward evaluated by expanding
  $\ds{1 \over 1 + \expo{-t}}$ in powers of $\ds{\expo{-t}}$. Expression \eqref{1} is reduced to:

\begin{align}
\mrm{S}\pars{n} & \equiv
{\pars{2n - 1}! \over \root{2}}\,\pars{4 \over \pi}^{2n}\sum_{k = 0}^{\infty}
{\pars{-1}^{k\pars{k + 1}/2} \over \pars{2k + 1}^{2n}}
\\[5mm] & =
\bbx{\ds{{\root{2} \over 2}\,{\Gamma\pars{2n} \over \pars{2\pi}^{2n}}\bracks{%
\zeta\pars{2n,{1 \over 8}} - \zeta\pars{2n,{5 \over 8}} -
\zeta\pars{2n,{3 \over 8}} + \zeta\pars{2n,{7 \over 8}}}}}
\end{align}
A: The above series is the expansion of the function Sin[x]/Cos[2x]
A: $$ \begin{align} 
\sum_{k=0}^{\infty}\frac{(-1)^{\frac{\large k(k+1)}{2}}}{(2k+1)^{2n}} &= \sum_{k=0}^{\infty}\left[\small\frac{(-1)^{\frac{\large (4k+0)(4k+1)}{2}}}{(2(4k+0)+1)^{2n}}+\frac{(-1)^{\frac{\large (4k+1)(4k+2)}{2}}}{(2(4k+1)+1)^{2n}}+\frac{(-1)^{\frac{\large (4k+2)(4k+3)}{2}}}{(2(4k+2)+1)^{2n}}+\frac{(-1)^{\frac{\large (4k+3)(4k+4)}{2}}}{(2(4k+3)+1)^{2n}}\normalsize\right] \\[3mm] 
&= \sum_{k=0}^{\infty}\left[\frac{1}{(8k+1)^{2n}}\color{red}{-}\frac{1}{(8k+3)^{2n}}\color{red}{-}\frac{1}{(8k+5)^{2n}}\color{red}{+}\frac{1}{(8k+7)^{2n}}\right] \\[3mm] 
&= \frac{1}{8^{2n}}\left[\zeta\left(2n,\,\frac18\right)-\zeta\left(2n,\,\frac38\right)-\zeta\left(2n,\,\frac58\right)+\zeta\left(2n,\,\frac78\right)\right] 
\end{align} $$ 
Hence, 
$$ 
S(n)=\color{red}{\frac{(2n-1)!}{\sqrt{2}\,(2\pi)^{2n}}\,\left[\zeta\left(2n,\,\frac18\right)-\zeta\left(2n,\,\frac38\right)-\zeta\left(2n,\,\frac58\right)+\zeta\left(2n,\,\frac78\right)\right]}
$$
  
 
It is also important to mention that it seems their are some relation with:
$$ \zeta(2n)= \frac{1}{(2n-1)!}\,\int_{0}^{\infty}\frac{x^{2n-1}}{e^x-1}\,dx = \frac{\left|{B}_{2n}\right|}{2}\,\frac{(2\pi)^{2n}}{(2n)!} $$ 
One would notice $\,\left(\frac18+\frac78\right)-\left(\frac38+\frac58\right)=\frac88-\frac88=0\,$: 
$$ \small \zeta\left(2n,\,\frac18\right)-\zeta\left(2n,\,\frac38\right)-\zeta\left(2n,\,\frac58\right)+\zeta\left(2n,\,\frac78\right) = \frac{1}{(2n-1)!}\,\int_{0}^{\infty}\frac{x^{2n-1}}{e^x-1}\left(e^{\frac78x}-e^{\frac58x}-e^{\frac38x}+e^{\frac18x}\right)dx $$
