Forma cerrada para las sumas $S(n)={(2n-1)!\over \sqrt2}\cdot{\left(4/ \pi\right)^{2n}}\cdot\sum\limits_{k=0}^{\infty}(-1)^{k(k+1)/2}(2k+1)^{-2n}$

Question

Considere las sumas

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

Tenemos $S(1)=1$ , $S(2)=11$ , $S(3)=361$ , $S(4)=24611$ .

No puedo detectar ningún patrón dentro de esta secuencia.

¿Cómo podemos elaborar una forma cerrada para $S(n)$ ?

Answer 1

Podemos observar que $$T(n)=\sum_{k\geq 0}\frac{(-1)^{k(k+1)/2}}{(2k+1)^{2n}} = \sum_{m\geq 1}\frac{\chi(m)}{m^{2n}} = L(\chi,2n)\tag{1}$$ es un Dirichlet $L$ -función asociada a la función multiplicativa $\chi(m)$ , que es igual a $0$ si $m$ está en paz, $1$ si $m\equiv \pm 1\pmod{8}$ y $-1$ si $m\equiv \pm 3\pmod{8}$ . En particular $$T(n) = \prod_{p>2}\left(1-\frac{\left(\frac{2}{p}\right)}{p^{2n}}\right)^{-1}\tag{2}$$ donde $\left(\frac{2}{p}\right)$ es el símbolo de Legendre. De la representación integral de Hazem Orabi $$S(n) = \frac{1}{\sqrt{2}(2\pi)^{2n}}\int_{0}^{+\infty}\frac{x^{2n-1}}{e^x-1}\left(e^{x/8}-e^{3x/8}-e^{5x/8}+e^{7x/8}\right)\,dx \tag{3}$$ también tenemos: $$S(n) = \frac{1}{\sqrt{2}}\left(\frac{4}{\pi}\right)^{2n}\int_{1}^{+\infty}\log(x)^{2n-1}\frac{x^2-1}{1+x^4}\,dx \tag{4}$$ o $$S(n) = \frac{1}{\sqrt{2}}\left(\frac{4}{\pi}\right)^{2n}\left.\frac{d^{2n-1}}{d\alpha^{2n-1}}\int_{1}^{+\infty}\frac{x^{2+\alpha}-x^{\alpha}}{1+x^4}\,dx\, \right|_{\alpha=0^+}\tag{5}$$ así que $S(n)$ depende de $\psi^{(2n-1)}(z)$ (el $(2n-1)$ -derivada de la función digamma, es decir, la $2n$ -derivada de $\log\Gamma$ ) evaluado en $z=\frac{1}{8},\frac{3}{8},\frac{5}{8},\frac{7}{8}$ . Por el fórmulas de reflexión para el $\psi^{(2n-1)}(z)$ función todo se reduce a las derivadas de la función $\cot(\pi z)$ en $z=\frac{1}{8}$ y $z=\frac{3}{8}$ . Esto demuestra la afirmación de G.H.Hardy:

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

En particular, $S(n)$ puede expresarse en términos de Polinomios de Euler $E_n(z)$ evaluado en $z=\frac{1}{4}$ y $z=\frac{3}{4}$ .

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $$\begin{align} \mrm{S}\pars{n} & \equiv {\pars{2n - 1}! \over \root{2}}\,\pars{4 \over \pi}^{2n}\sum_{k = 0}^{\infty} {\pars{-1}^{k\pars{k + 1}/2} \over \pars{2k + 1}^{2n}} \\[5mm] & = \pars{2n - 1}!\,{2^{4n - 1/2} \over \pi^{2n}}\bracks{% \sum_{k = 0}^{\infty}{\pars{-1}^{k} \over \pars{4k + 1}^{2n}} + \sum_{k = 0}^{\infty}{\pars{-1}^{k + 1} \over \pars{4k + 3}^{2n}}} \\[5mm] & = {\root{2} \over 2\pi^{2n}}\bracks{% \pars{2n - 1}!\sum_{k = 0}^{\infty}{\pars{-1}^{k} \over \pars{k + 1/4}^{2n}} - \pars{2n - 1}!\,\sum_{k = 0}^{\infty}{\pars{-1}^{k} \over \pars{k + 3/4}^{2n}}} \label{1}\tag{1} \end{align}$$

Sin embargo,

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

La última integral, en $\eqref{2}$, se evalúa directamente expandiendo $\ds{1 \over 1 + \expo{-t}}$ en los poderes de $\ds{\expo{-t}}$ . La expresión $\eqref{1}$ se reduce a:

$$\begin{align} \mrm{S}\pars{n} & \equiv {\pars{2n - 1}! \over \root{2}}\,\pars{4 \over \pi}^{2n}\sum_{k = 0}^{\infty} {\pars{-1}^{k\pars{k + 1}/2} \over \pars{2k + 1}^{2n}} \\[5mm] & = \bbx{\ds{{\root{2} \over 2}\,{\Gamma\pars{2n} \over \pars{2\pi}^{2n}}\bracks{% \zeta\pars{2n,{1 \over 8}} - \zeta\pars{2n,{5 \over 8}} - \zeta\pars{2n,{3 \over 8}} + \zeta\pars{2n,{7 \over 8}}}}} \end{align}$$

Answer 3

$$\begin{align} \sum_{k=0}^{\infty}\frac{(-1)^{\frac{\large k(k+1)}{2}}}{(2k+1)^{2n}} &= \sum_{k=0}^{\infty}\left[\small\frac{(-1)^{\frac{\large (4k+0)(4k+1)}{2}}}{(2(4k+0)+1)^{2n}}+\frac{(-1)^{\frac{\large (4k+1)(4k+2)}{2}}}{(2(4k+1)+1)^{2n}}+\frac{(-1)^{\frac{\large (4k+2)(4k+3)}{2}}}{(2(4k+2)+1)^{2n}}+\frac{(-1)^{\frac{\large (4k+3)(4k+4)}{2}}}{(2(4k+3)+1)^{2n}}\normalsize\right] \\[3mm] &= \sum_{k=0}^{\infty}\left[\frac{1}{(8k+1)^{2n}}\color{red}{-}\frac{1}{(8k+3)^{2n}}\color{red}{-}\frac{1}{(8k+5)^{2n}}\color{red}{+}\frac{1}{(8k+7)^{2n}}\right] \\[3mm] &= \frac{1}{8^{2n}}\left[\zeta\left(2n,\,\frac18\right)-\zeta\left(2n,\,\frac38\right)-\zeta\left(2n,\,\frac58\right)+\zeta\left(2n,\,\frac78\right)\right] \end{align}$$ Por lo tanto, $$S(n)=\color{red}{\frac{(2n-1)!}{\sqrt{2}\,(2\pi)^{2n}}\,\left[\zeta\left(2n,\,\frac18\right)-\zeta\left(2n,\,\frac38\right)-\zeta\left(2n,\,\frac58\right)+\zeta\left(2n,\,\frac78\right)\right]}$$

También es importante mencionar que parece que hay alguna relación con: $$\zeta(2n)= \frac{1}{(2n-1)!}\,\int_{0}^{\infty}\frac{x^{2n-1}}{e^x-1}\,dx = \frac{\left|{B}_{2n}\right|}{2}\,\frac{(2\pi)^{2n}}{(2n)!}$$ Uno se daría cuenta de que $\,\left(\frac18+\frac78\right)-\left(\frac38+\frac58\right)=\frac88-\frac88=0\,$ : $$\small \zeta\left(2n,\,\frac18\right)-\zeta\left(2n,\,\frac38\right)-\zeta\left(2n,\,\frac58\right)+\zeta\left(2n,\,\frac78\right) = \frac{1}{(2n-1)!}\,\int_{0}^{\infty}\frac{x^{2n-1}}{e^x-1}\left(e^{\frac78x}-e^{\frac58x}-e^{\frac38x}+e^{\frac18x}\right)dx$$

Answer 4

La serie anterior es la expansión de la función Sin[x]/Cos[2x]