Evaluar:: $ 2 \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n+1}\left( 1 + \frac12 +\cdots + \frac 1n\right) $

Question

Cómo evaluar la serie: $$ 2 \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n+1}\left( 1 + \frac12 + \cdots + \frac 1n\right) $$

Según Mathematica, esto converge a $ (\log 2)^2 $ .

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &2\sum_{n = 1}^{\infty}{\pars{-1}^{n + 1} \over n + 1} \pars{1 + {1 \over 2} + \cdots + {1 \over n}} = -2\sum_{n = 1}^{\infty}\pars{-1}^{n}\, H_{n}\int_{0}^{1}x^{n}\,\dd x = -2\int_{0}^{1}\sum_{n = 1}^{\infty}H_{n}\pars{-x}^{n}\,\dd x \\[5mm] = &\ -2\int_{0}^{1}\braces{-\,{\ln\pars{1 - \bracks{-x}} \over 1 - \pars{-x}}}\,\dd x = 2\int_{0}^{1}{\ln\pars{1 +x} \over 1 + x}\,\dd x = \left.\ln^{2}\pars{1 + x}\,\right\vert_{\ x\ =\ 0}^{\ x\ =\ 1} = \bbx{\ds{\ln^{2}\pars{2}}} \end{align}