Cómo evaluar $ \sum\limits_{n=1}^{\infty}\frac{1}{n}\left(\frac{1}{n+1}-\frac{1}{n+2}+\frac{1}{n+3}-\cdots\right)^{2}$

Question

Cómo probar $$ \sum\limits_{n=1}^{\infty}\frac{1}{n}\left(\frac{1}{n+1}-\frac{1}{n+2}+\frac{1}{n+3}-\cdots\right)^{2}=\frac{\pi^2\ln2}{6}-\frac{\ln^32}{3}-\frac{3\zeta(3)}{4} $$ $\mathbf {My\,Attempt:}$
Puse $$\left(\frac{1}{n+1}-\frac{1}{n+2}+\frac{1}{n+3}-\cdots\right) = \int_0^1 \frac{x^n}{1+x}dx$$ y por lo tanto
$$\left(\frac{1}{n+1}-\frac{1}{n+2}+\frac{1}{n+3}-\cdots\right)^2=\int_0^1\int_0^1 \frac{(xy)^n}{(1+x)(1+y)}dxdy$$
Así, la suma es igual a $$\int_0^1\int_0^1 \frac{1}{(1+x)(1+y)}\left(\sum\limits_{n=1}^{\infty} \frac{(xy)^n}{n}\right)dxdy=-\int_0^1\int_0^1 \frac{\ln(1-xy)}{(1+x)(1+y)}dxdy$$ La integral interna es $$\int_0^1 \frac{\ln(1-xy)}{1+y}dy=\mathrm{Li}_2 \left(\frac{1-x}{1+x}\right)-\mathrm{Li}_2\left(\frac{1}{1+x}\right)+\ln(1-x)\ln\left(\frac{2x}{1+x}\right)$$ El resto es un montón de cálculos.

$\text{Any hint for a better method or idea?}$

Answer 1

La segunda ronda de integración se completa con el uso de \begin {align} I_{1} &= \int_ {0}^{1} \mathrm {Li}_2 \left ( \frac {1-x}{1+x} \right ) \, \frac {dx}{1+x} = \zeta (2) \, \ln 2 - \frac {5}{8} \, \zeta (3) \\ I_{2} &= \int_ {0}^{1} \mathrm {Li}_2 \left ( \frac {1}{1+x} \right ) \, \frac {dx}{1+x} = \frac { \zeta (3)}{8} + \frac { \zeta (2) \, \ln 2}{2} - \frac { \ln ^{3}2}{6} \\ I_{3} &= \int_ {0}^{1} \ln (1-x) \, \ln\left ( \frac {2 x}{1+x} \right ) \, \frac {dx}{1+x} = \frac {3 \, \zeta (3)}{2} + \frac { \ln ^{3}2}{6} - \frac {3}{2} \, \zeta (2) \, \ln 2. \end {align}

Con estos valores entonces: \begin {align} S &= \sum_ {n=1}^{ \infty } \frac {1}{n} \, \left ( \sum_ {k=1}^{ \infty } \frac {(-1)^{k-1}}{k+n} \right )^{2} \\ &= \sum_ {n=1}^{ \infty } \frac {1}{n} \, \int_ {0}^{1} \int_ {0}^{1} \frac {(u t)^{n} \N, du \N, dt}{(1+u) (1+t)} \\ &= - \int_ {0}^{1} \int_ {0}^{1} \frac { \ln (1- u t) \N, du \N, dt}{(1+u)(1+t)} \\ &= - \int_ {0}^{1} \left ( \mathrm {Li}_2 \left ( \frac {1-t}{1+t} \right )- \mathrm {Li}_2 \left ( \frac {1}{1+t} \right )+ \ln (1-t) \ln\left ( \frac {2t}{1+t} \right ) \right ) \, \frac {dt}{1+t} \\ &= - \left [ \left ( \zeta (2) \, \ln 2 - \frac {5}{8} \, \zeta (3) \right ) - \left ( \frac { \zeta (3)}{8} + \frac { \zeta (2) \, \ln 2}{2} - \frac { \ln ^{3}2}{6} \right ) \right. \\ & \hspace {20mm} \left. + \left ( \frac {3 \, \zeta (3)}{2} + \frac { \ln ^{3}2}{6} - \frac {3}{2} \, \zeta (2) \, \ln 2 \right ) \right ] \\ &= \zeta (2) \, \ln 2 - \frac {3 \, \zeta (3)}{4} - \frac { \ln ^{3}2}{3}. \end {align}

Esto lleva a decir $$ \sum\limits_{n=1}^{\infty}\frac{1}{n}\left(\frac{1}{n+1}-\frac{1}{n+2}+\frac{1}{n+3}-\cdots\right)^{2}= \zeta(2)\ln 2 -\frac{3\zeta(3)}{4} - \frac{\ln^{3}2}{3}.$$

Answer 2

$$\iint_{(0,1)^2}\frac{-\log(1-xy)}{(1+x)(1+y)}\,dx\,dy = 2\iint_{0\leq y\leq x\leq 1}\frac{-\log(1-xy)}{(1+x)(1+y)}\,dx\,dy $$ es igual a $$ 2\int_{0}^{1}\int_{0}^{1}\frac{-x\log(1-x^2 z)}{(1+x)(1+xz)}\,dx\,dz = 2\int_{0}^{1}\frac{\text{Li}_2\left(\frac{1}{1+x}\right)-\text{Li}_2(1-x)-\log(x)\log(1-x^2)}{1+x}\,dx$$ y podemos abordar tres integrales distintas. $$ \int_{0}^{1}\text{Li}_2\left(\frac{1}{1+x}\right)\frac{dx}{1+x}=\int_{1/2}^{1}\frac{\text{Li}_2(x)}{x}\,dx=\int_{1/2}^{1}\sum_{n\geq 1}\frac{x^{n-1}}{n^2}\,dx=\zeta(3)-\text{Li}_3\left(\tfrac{1}{2}\right), $$ $$ \int_{0}^{1}\frac{\text{Li}_2(1-x)}{1+x}\,dx = \int_{0}^{1}\frac{\text{Li}_2(x)}{2-x}\,dx\stackrel{\text{IBP}}{=}-\int_{0}^{1}\frac{\log(1-x)\log(2-x)}{x}\,dx=\frac{\pi^2}{4}\log(2)-\zeta(3)$$ y la tercera es similar. Curiosamente, la segunda integral es igual a $$ \sum_{n\geq 1}\frac{1}{n}\left[\frac{1}{(n+1)^2}+\ldots+\frac{1}{(2n)^2}\right]=\frac{\pi^2}{4}\log(2)-\zeta(3).$$