Prueba $\sum_{k=1}^{\infty }\frac{\left ( -1 \right )^{k}}{k}\sum_{j=1}^{2k}\frac{\left ( -1 \right )^{j}}{j}=\frac{\pi ^{2}}{48}+\frac{1}{4}\ln^22$ .

Question

Cómo probar la siguiente serie, $$\sum_{k=1}^{\infty }\frac{\left ( -1 \right )^{k}}{k}\sum_{j=1}^{2k}\frac{\left ( -1 \right )^{j}}{j}=\frac{\pi ^{2}}{48}+\frac{1}{4}\ln^22$$ Conozco una fórmula que podría ser útil. $$\sum_{j=1}^{n}\frac{\left ( -1 \right )^{j-1}}{j}=\ln 2+\left ( -1 \right )^{n-1}\int_{0}^{1}\frac{x^{n}}{1+x}\mathrm{d}x$$ cualquier sugerencia será apreciada.

Answer 1

Sí. Como has mencionado, es una fórmula muy útil.

Utilice $$\sum_{j=1}^{n}\frac{\left ( -1 \right )^{j-1}}{j}=\ln 2+\left ( -1 \right )^{n-1}\int_{0}^{1}\frac{x^{n}}{1+x}\, \mathrm{d}x$$ obtenemos \begin{align*} &\sum_{k=1}^{\infty }\frac{\left ( -1 \right )^{k}}{k}\sum_{j=1}^{2k}\frac{\left ( -1 \right )^{j}}{j}=\ln 2\sum_{k=1}^{\infty }\frac{\left ( -1 \right )^{k}}{k}-\sum_{k=1}^{\infty }\frac{\left ( -1 \right )^{k}}{k}\int_{0}^{1}\frac{x^{2k}}{1+x}\, \mathrm{d}x\\ &=\ln^22+\int_{0}^{1}\frac{1}{1+x}\sum_{k=1}^{\infty }\frac{\left ( -x^{2} \right )^{k}}{k}\, \mathrm{d}x=\ln^22-\int_{0}^{1}\frac{\ln\left ( 1+x^{2} \right )}{1+x}\, \mathrm{d}x \end{align*} dejar $$f\left ( t \right )=\int_{0}^{1}\frac{\ln\left ( 1+tx^{2} \right )}{1+x}\,\mathrm{d}x$$ entonces \begin{align*} f{}'\left ( t \right ) &=\int_{0}^{1}\frac{x^{2}}{\left ( 1+x \right )\left ( 1+tx^{2} \right )}\,\mathrm{d}x \\ &=\frac{1}{t+1}\int_{0}^{1}\frac{x-1}{1+tx^{2}}\,\mathrm{d}x+\frac{1}{t+1}\int_{0}^{1}\frac{\mathrm{d}x}{x+1} \\ &=\frac{1}{t+1}\left [ \frac{1}{2t}\ln\left ( 1+tx^{2} \right )-\frac{1}{\sqrt{t}}\arctan\left ( \sqrt{t}x \right )+\ln\left ( x+1 \right ) \right ]_{0}^{1} \\ &=\frac{1}{t+1}\left [ \frac{1}{2t}\ln\left ( 1+t \right )-\frac{1}{\sqrt{t}}\arctan\left ( \sqrt{t} \right )+\ln 2 \right ] \end{align*} Integrar la espalda $$\Rightarrow f\left ( t \right )=\frac{1}{2}\left [ -\mathrm{Li}_{2}\left ( -t \right )-\frac{1}{2}\ln^{2}\left ( t+1 \right ) \right ]-\arctan^{2}\sqrt{t}+\ln 2\ln\left ( t+1 \right )$$ entonces deja que $t=1$ y utilizar $\displaystyle \mathrm{Li}_{2}\left ( -1 \right )=-\frac{\pi ^{2}}{12}$ , obtendrá la respuesta deseada.

EDITAR: $$\mathrm{Li}_{2}\left ( -1 \right )=\sum_{k=1}^{\infty }\frac{\left ( -1 \right )^{k}}{k^{2}}=\sum_{k=1}^{\infty }\frac{1}{k^{2}}-2\sum_{k=1}^{\infty }\frac{1}{\left ( 2n-1 \right )^{2}}=\frac{\pi ^{2}}{6}-2\cdot \frac{\pi ^{2}}{8}=-\frac{\pi ^{2}}{12}$$

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\sum_{k = 1}^{\infty }{\pars{-1}^{k} \over k} \sum_{j = 1}^{2k}{\pars{-1}^{\,j} \over j}:\ {\large ?}}$ .

Tenga en cuenta que \begin{align} &\sum_{j = 1}^{2k}{\pars{-1}^{\,j} \over j} = \sum_{j = 1}^{k}{1 \over 2j} - \sum_{j = 1}^{k}{1 \over 2j - 1} = {1 \over 2}\sum_{j = 1}^{k}{1 \over j} - \pars{\sum_{j = 1}^{2k}{1 \over j} - \sum_{j = 1}^{k}{1 \over 2j}} = \sum_{j = 1}^{k}{1 \over j} - \sum_{j = 1}^{2k}{1 \over j} \\[5mm] = &\ \bbx{\ds{H_{k} - H_{2k}}}\qquad \pars{~H_{n}:\ Harmonic\ Number~} \end{align}

---

Entonces, \begin{align} &\sum_{k = 1}^{\infty }{\pars{-1}^{k} \over k} \sum_{j = 1}^{2k}{\pars{-1}^{\,j} \over j} = \sum_{k = 1}^{\infty }{\pars{-1}^{k} \over k}\pars{H_{k} - H_{2k}} = \sum_{k = 1}^{\infty }{\pars{-1}^{k} \over k}H_{k} - 2\sum_{k = 1}^{\infty }{\ic^{2k} \over 2k}H_{2k} \\[5mm] = &\ \sum_{k = 1}^{\infty }{\pars{-1}^{k} \over k}H_{k} - 2\sum_{k = 1}^{\infty }{\ic^{k} \over k}H_{k}\,{1 + \pars{-1}^{k} \over 2} = \sum_{k = 1}^{\infty }{\pars{-1}^{k} \over k}H_{k} - 2\,\Re\sum_{k = 1}^{\infty }{\ic^{k} \over k}H_{k} \\[5mm] = &\ \,\mrm{f}\pars{-1} - 2\,\Re\pars{\mrm{f}\pars{\ic}}\quad \mbox{where}\quad\mrm{f}\pars{x} \equiv \sum_{k = 1}^{\infty}{x^{k} \over k}\,H_{k}\label{1}\tag{1} \end{align}

Sin embargo, \begin{align} \mrm{f}\pars{x} & = \sum_{k = 1}^{\infty}x^{k}\,H_{k}\int_{0}^{1}t^{k - 1}\,\dd t = \int_{0}^{1}{1 \over t}\sum_{k = 1}^{\infty}\pars{xt}^{k}\,H_{k}\,\dd t = \int_{0}^{1}{1 \over t}\bracks{-\,{\ln\pars{1 - xt} \over 1 - xt}}\,\dd t \\[5mm] & = -\int_{0}^{x}{\ln\pars{1 - t} \over t\pars{1 - t}}\,\dd t = -\int_{0}^{x}{\ln\pars{1 - t} \over t}\,\dd t - \int_{0}^{x}{\ln\pars{1 - t} \over 1 - t}\,\dd t \\[5mm] & = \int_{0}^{x}\mrm{Li}_{2}'\pars{t}\,\dd t + \bracks{{1 \over 2}\,\ln^{2}\pars{1 - t}}_{\ 0}^{\ x} = \bbx{\ds{\mrm{Li}_{2}\pars{x}} + {1 \over 2}\,\ln^{2}\pars{1 - x}} \end{align}

Con la expresión \eqref{1}: \begin{align} &\sum_{k = 1}^{\infty }{\pars{-1}^{k} \over k} \sum_{j = 1}^{2k}{\pars{-1}^{\,j} \over j} = \mrm{Li}_{2}\pars{-1} + {1 \over 2}\,\ln^{2}\pars{2} -2\,\Re\pars{\mrm{Li}_{2}\pars{\ic}} - \Re\pars{\ln^{2}\pars{1 - \ic}} \\[5mm] = & \bbx{\ds{{\pi^{2} \over 48} + {1 \over 4}\,\ln^{2}\pars{2}}} \end{align}

Tenga en cuenta que $\ds{\mrm{Li}_{2}\pars{-1} = -\,{\pi^{2} \over 12}}$ , $\ds{\Re\pars{\mrm{Li}_{2}\pars{\ic}} = -\,{\pi^{2} \over 48}}$ y $$ \Re\pars{\ln^{2}\pars{1 - \ic}} = \Re\pars{\bracks{{1 \over 2}\,\ln\pars{2} - {\pi \over 4}\,\ic}^{2}} = {1 \over 4}\,\ln^{2}\pars{2} - {\pi^{2} \over 16} $$