Demostrar que $\ln{\left({A^6\sqrt{\pi}\over 2^{7\over6}e}\right)}=\sum\limits_{n=1}^{\infty}{(-1)^{n+1}\over n(n+1)}\eta(n)$

Question

Mostrar que

$$\ln{\left({A^6\sqrt{\pi}\over 2^{7\over6}e}\right)}=\sum_{n=1}^{\infty}{(-1)^{n+1}\over n(n+1)}\eta(n)$$

(donde $\eta(n)$ es la de Dirichlet eta función, y Una es la Glaisher-Kinkelin constante).

Trato:

$$\ln{\left({A^6\sqrt{\pi}\over 2^{7\over6}e}\right)}=\sum_{n=1}^{\infty}{(-1)^{n+1}\over n}\eta(n)-\sum_{n=1}^{\infty}{(-1)^{n+1}\over n+1}\eta(n)$$

Utilizamos esta serie $$\ln{{\pi\over 2}}=\sum_{n=1}^{\infty}{(-1)^{n+1}\over n}\eta(n)$$ a simplificar a

$$-\ln{\left({A^6\over 2^{1\over6}\sqrt{\pi}e}\right)}=\sum_{n=1}^{\infty}{(-1)^{n+1}\over n+1}\eta(n)$$

Tenemos

$$\ln{\left[\prod_{k=1}^{\infty}\left({k\over k+1}\right)^{(-1)^{k+1}}\right]}=\sum_{n=1}^{\infty}{(-1)^n\eta(n)\over n}$$

No podemos usar esto para aplicar en la anterior serie.

Me pregunto, ¿hay un infinito de productos para

$$\ln{\left[F(k)\right]}=\sum_{n=1}^{\infty}{(-1)^n\eta(n)\over 1+n}$$

¿Alguien puede por favor dar una mano aquí? Gracias.

Answer 1

Sugerencia. Uno puede escribir $$ \begin{align} \sum_{n=1}^{\infty}{(-1)^{n+1}\over n+1}\eta(n)&=\sum_{n=1}^{\infty}{(-1)^{n+1}\over n+1}\eta(n) \\&=\sum_{n=1}^{\infty}{(-1)^{n+1}\over n+1}\left(1+\sum_{k=2}^{\infty}\frac{(-1)^{k-1}}{k^n}\right) \\&=\sum_{n=1}^{\infty}{(-1)^{n+1}\over n+1}+\sum_{n=1}^{\infty}{(-1)^{n+1}\over n+1}\sum_{k=2}^{\infty}\frac{(-1)^{k-1}}{k^n} \\&=1-\ln 2+\sum_{k=2}^{\infty}(-1)^{k-1}\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{(n+1)k^n} \\&=1-\ln 2+\sum_{k=2}^{\infty}(-1)^{k-1}\left[1-k\ln\left(1+\frac1k \right) \right] \\&=\sum_{k=1}^{\infty}(-1)^{k}\left[k\ln\left(1+\frac1k \right)-1 \right] \tag1 \end{align} $$ since, by a Taylor series expansion, as $k\to \infty$, uno tiene $$ (-1)^{k}\left[k\ln\left(1+\frac1k \right)-1 \right]=\frac{(-1)^{k}}{2k}+O\left(\frac{1}{k^2}\right) $$ the series in $(1)$ es por lo tanto convergente, entonces es suficiente para considerar su suma parcial $$ \begin{align} S_{2n}&=\sum_{k=1}^{2n}(-1)^{k}\left[k\ln\left(1+\frac1k \right)-1 \right] \\&=\sum_{k=1}^{2n}(-1)^{k}k\ln\left(k+1\right)-\sum_{k=1}^{2n}(-1)^{k}k\ln k-\sum_{k=1}^{2n}(-1)^{k} \\&=\sum_{k=2}^{2n+1}(-1)^{k-1}(k-1)\ln k-\sum_{k=1}^{2n}(-1)^{k}k\ln k \\&=2n\ln (2n+1)-2\sum_{k=1}^{2n}(-1)^{k}k\ln k+\sum_{k=1}^{2n}(-1)^{k}\ln k \tag2 \end{align} $ de$ la última suma se obtiene a través de la fórmula de Stirling $$ \sum_{k=1}^{2n}(-1)^{k}\ln k=\frac{\ln \pi}2+\frac{\ln n}2+O\left(\frac1n \right) $$ y uno puede reorganizar la suma $$ \begin{align} &\sum_{k=1}^{2n}(-1)^{k-1}k\ln k \\&=\sum_{k=1}^{2n}k\ln k-2\sum_{k=1}^{n}2k\ln(2k) \\&=\sum_{k=1}^{2n}k\ln k-4\sum_{k=1}^{n}k\ln k-4\sum_{k=1}^{n}k\ln 2 \\&=\ln \frac{1^12^2\cdots (2n)^{2n}}{(2n)^{2n^2+n+1/12}e^{-n^2}}-4\ln \frac{1^12^2\cdots n^n}{n^{n^2/2+n/2+1/12}e^{-n^2/4}}-\left(n-\frac1{12}\right)\ln 2-\left(n+\frac14\right)\ln n \end{align} $$ giving in $(2)$ $$ S_{2n}=\frac{\ln \pi}2+\frac{\ln 2}6+1-6\ln A+O\left(\frac1n \right) $$ como se esperaba, ya que $$ A= \lim_{n \to \infty}\frac{1^12^2\cdots n^n}{n^{n^2/2+n/2+1/12}e^{-n^2/4}}. $$

Answer 2

El uso de la generalización $Q_m(x)$ de la función Gamma en

https://www.fernuni-hagen.de/analysis/download/bachelorarbeit_aschauer.pdf

con $\enspace\displaystyle \ln Q_m(x)=\frac{(-x)^{m+1}}{m+1}\gamma +\sum\limits_{n=2}^\infty\frac{(-x)^{m+n}}{m+n}\zeta(n)\enspace$ en la página $13$, $(4.2)$con

$Q_0(1)=\Gamma(2)=1\enspace$ $\enspace\displaystyle Q_0(\frac{1}{2})=\Gamma(\frac{3}{2})=\frac{\sqrt{\pi}}{2}\enspace$ y con los valores calculados

$\displaystyle Q_1:=Q_1(1)=\frac{\sqrt{2\pi}}{e}\enspace$ $\enspace\displaystyle Q_1(\frac{1}{2})= A_1^{\frac{3}{2}}2^{\frac{5}{24}}e^{-\frac{1}{2}} \enspace$ page $38$ $\enspace$ donde

$A_1\equiv A\enspace$ (Glaisher Kinkelin constante), llegamos a la

$\displaystyle \sum\limits_{n=1}^\infty \frac{(-1)^{n+1}}{n(n+1)}\eta(n)=$

$\displaystyle = \frac{\eta(1)}{2}-\sum\limits_{n=2}^\infty\frac{(-1)^n}{n}\zeta(n)+2\sum\limits_{n=2}^\infty\frac{(-1)^n}{2^n n}\zeta(n)-\sum\limits_{n=2}^\infty\frac{(-1)^{n+1}}{n+1}\zeta(n)+4\sum\limits_{n=2}^\infty\frac{(-1)^{n+1}}{2^{n+1}(n+1)}\zeta(n)$

$\displaystyle =\frac{\eta(1)}{2}-( \ln Q_0(1)+\gamma )+2( \ln Q_0(\frac{1}{2})+\frac{\gamma}{2} )-( \ln Q_1(1)-\frac{\gamma}{2})+4(\ln Q_1(\frac{1}{2})-\frac{\gamma}{8})$

$\displaystyle =\frac{\ln 2}{2}+2 \ln\frac{\sqrt{\pi}}{2} -\ln\frac{\sqrt{2\pi}}{e} +4\ln(A_1^{\frac{3}{2}}2^{\frac{5}{24}}e^{-\frac{1}{2}})=\ln\frac{A^6\sqrt{\pi}}{2^{7/6}e}$

Answer 3

Otro enfoque es el uso común de representación integral de Dirichlet eta función, junto con una representación integral de la zeta de Hurwitz función.

En concreto, se puede utilizar $$\eta(s) = \frac{1}{\Gamma(s)}\int_{0}^{\infty} \frac{x^{s-1}}{1+e^{x}} \, dx , \quad \text{Re}(s) >0,\tag{1}$$ and $$\zeta(s,z) = \frac{1}{\Gamma(s)}\int_{0}^{\infty} \frac{x^{s-1}e^{-zx}}{1-e^{-x}} \, dx \tag{2}, \quad (\text{Re}(s) >1, \, \text{Re}(z) >0). $$

De $(2)$ se sigue que

$$ \begin{align} \zeta(s)-\zeta \left(s, \frac{3}{2} \right) &= \frac{1}{\Gamma(s)} \int_{0}^{\infty}\frac{x^{s-1}}{1-e^{-x}} \left(e^{-x}-e^{-(3/2)x} \right) \\ &= \frac{2^{s}}{\Gamma(s)} \int_{0}^{\infty} \frac{u^{s-1}}{1-e^{-2u}} \left(e^{-2u} - e^{-3u} \right) \, du \\ &= \frac{2^{s}}{\Gamma(s)} \int_{0}^{\infty} \frac{u^{s-1}e^{-2u}} {1+e^{-u}} \, du \\ &= \frac{2^{s}}{\Gamma(s)} \int_{0}^{\infty} \frac{u^{s-1}e^{-u}} {1+e^{u}} \, du, \quad \text{Re}(s) >0. \end{align}$$

Y el uso de $(1)$, obtenemos $$ \begin{align} &\sum_{n=1}^{\infty} \frac{(-1)^{n+1} }{n(n+1)} \, \eta(n) \\ &= \sum_{n=1}^{\infty}\frac{(-1)^{n+1} }{n(n+1)}\frac{1}{\Gamma(n)} \int_{0}^{\infty} \frac{x^{n-1}}{1+e^{x}} \, dx \\ &= \int_{0}^{\infty} \frac{1}{1+e^{x}} \frac{1}{x^{2}} \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n!} \, \frac{x^{n+1}}{n+1} \, dx \\ &= \int_{0}^{\infty} \frac{1}{1+e^{x}} \frac{1}{x^{2}} \left(e^{-x}+x-1 \right) \, dx \\ &= \lim_{s \to -1^{+}} \int_{0}^{\infty} \frac{x^{s-1}}{1+e^{x}} \left(e^{-x}+x-1 \right) \, dx \\ &= \lim_{s \to -1^{+}} \left[2^{-s} \, \Gamma(s) \left( \zeta(s) - \zeta \left(s, \frac{3}{2} \right)\right) + \Gamma(s+1) \eta(s+1 ) - \Gamma(s) \eta(s) \right] \\&= \lim_{s \to -1^{+}} \left[2^{-s} \, \Gamma(s) \left(\zeta(s) - \zeta \left(s, \frac{1}{2} \right) +2^{s}\right) + \Gamma(s+1) \eta(s+1 ) - \Gamma(s) \eta(s) \right] \tag{3} \\&= \lim_{s \to -1^{+}} \left[2^{-s} \, \Gamma(s) \left(\zeta(s) - \left(2^{s}-1\right)\zeta (s) +2^{s}\right) + \Gamma(s+1) \eta(s+1 ) - \Gamma(s) \eta(s) \right] \tag{4} \\ &= \lim_{s \to -1^{+}} \left[\left(2^{1-s}-1 \right) \Gamma(s) \zeta(s) + \Gamma(s) + \Gamma(s+1) \eta(s+1 ) - \Gamma(s) \eta(s)\right] \\&= \lim_{s \to -1^{+}} \left[ \Gamma(s) + \Gamma(s+1) \eta(s+1) -2\Gamma(s) \eta(s) \right] \tag{5}. \end{align}$$

La expansión de alrededor de $s=-1$, obtenemos

$$ \begin{align} &\lim_{s \to -1^{+}} \left[ \Gamma(s) + \Gamma(s+1) \eta(s+1) -2\Gamma(s) \eta(s) \right] \\ &= \small \lim_{s \to -1^{+}} \Big[\left(- \frac{1}{s+1} + \gamma-1 + \mathcal{O}(s+1) \right) + \left(\frac{1}{s+1} - \gamma + \mathcal{O}(s+1) \right) \Big(\frac{1}{2} +\eta'(0)(s+1) \\ &+ \small \mathcal{O}\left( (s+1)^{2} \right) \Big) - \small 2 \left(- \frac{1}{s+1} + \gamma-1 + \mathcal{O}(s+1) \right) \left(\frac{1}{4} + \eta'(-1)(s+1) + \mathcal{O}\left((s+1)^{2} \right) \right) \Big] \\ &= \lim_{s \to -1^{+}} \left[\eta'(0) + 2 \eta'(-1) - \frac{1}{2} + \mathcal{O}\left((s+1) \right) \right] \\&= \frac{1}{2} \, \ln \left(\frac{\pi}{2} \right) +2 \big( 4 \ln(2) \zeta(-1) -3 \zeta'(-1) \big) -\frac{1}{2} \tag{6} \\&= \frac{1}{2} \, \ln \left(\frac{\pi}{2} \right) +2 \left(-\frac{1}{3} \, \ln(2) -3 \left(\frac{1}{12} - \ln(A) \right) \right) - \frac{1}{2} \tag{7} \\&= \ln \left(\frac{\sqrt{\pi} \, A^{6}}{ 2^{7/6} e} \right) . \end{align}$$

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$(3)$ $\zeta(s,a) = \zeta(s,a+1) + a^{-s}$

$(4)$ http://mathworld.wolfram.com/HurwitzZetaFunction.html (11)

$(5)$ $\eta(s) = (1-2^{1-s}) \zeta(s)$

$(6)$ http://mathworld.wolfram.com/DirichletEtaFunction.html (11)

$(7)$ https://en.wikipedia.org/wiki/Glaisher%E2%80%93Kinkelin_constant

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Expandir $\Gamma(s+1)$ en una de la serie de Laurent alrededor de $s=-1$, la utilización de la identidad de la $\Gamma(s+1) = \frac{\Gamma(s+2)}{s+1}$.