El valor de la presente

Question

¿Cómo calcular la siguiente serie convergente? ¡o alguna sugerencia! $$\sum_{n=1}^{\infty}(-1)^{n-1}\frac{\log n}{n}.$ $ El MATHEMATICA9.0 Wolfram da el resultado es $1/2(\log 2)^2-\gamma\log 2$, donde $\gamma$ tha constante de Euler.

Answer 1

Podemos escribir

$$\begin{align} \sum_{n=1}^{2N}(-1)^{n-1}\frac{\log(n)}{n}&=\sum_{n=1}^{2N} \frac{\log(n)}{n}-2\sum_{n=1}^N\frac{\log(2n)}{2n}\\\\ &=\sum_{n=N+1}^{2N}\frac{\log(n)}{n}-\log(2)\sum_{n=1}^N \frac1n \end {Alinee el} $$

Usando la fórmula de Euler-Maclaurin suma, tenemos

$$\begin{align} \sum_{n=}^{2N}(-1)^{n-1}\frac{\log(n)}{n}&=\int_{N+1}^{2N}\frac{\log(x)}{x}\,dx-\log(2)\,H_n+O\left(\frac{\log(N)}{N}\right)\\\\ &=\frac12\log^2(2N)-\frac12\log^2(N+1)-\log(2)\log(N)-\log(2)\gamma+O\left(\frac{\log(N)}{N}\right)\\\\ &=\frac12 \log^2(2)-\log(2)\gamma+O\left(\frac{\log(N)}{N}\right)\\\\ \end {Alinee el} $$

Tomando el límite como $N\to \infty$ produce el codiciado límite

$$\sum_{n=1}^{\infty}(-1)^{n-1}\frac{\log(n)}{n}=\frac12\log^2(2)-\log(2)\gamma$$

Answer 2

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

$\ds{\sum_{n = 1}^{\infty}\pars{-1}^{n - 1}\,{\ln\pars{n} \over n}:\ {\large ?}}$.

De ahora en adelante, voy a usar bien conocido identidades que están relacionados con la Riemann Zeta Función: \begin{align} &\sum_{n = 1}^{\infty}\pars{-1}^{n - 1}\,{\ln\pars{n} \over n} = \left.\partiald{}{\mu}\sum_{n = 1}^{\infty}{\pars{-1}^{n + 1} \over n^{1 - \mu}} \right\vert_{\ \mu\ =\ 0^{+}} = \left.\partiald{\bracks{\pars{1 - 2^{\mu}}\zeta\pars{1 - \mu}}}{\mu}\, \right\vert_{\ \mu\ =\ 0^{+}} \\[5mm] = &\ \partiald{}{\mu} \braces{\bracks{-\ln\pars{2}\mu - {1 \over 2}\,\ln^{2}\pars{2}\,\mu^{2} + \,\mrm{O}\pars{\mu^{3}}}\bracks{-\,{1 \over \mu} + \gamma + \,\mrm{O}\pars{\mu^{1}}}}_{\ \mu\ =\ 0^{+}} \\[5mm] = &\ \partiald{}{\mu}\braces{\ln\pars{2} + \bracks{{1 \over 2}\,\ln^{2}\pars{2} - \gamma\ln\pars{2}}\mu + \,\mrm{O}\pars{\mu^{2}}}_{\ \mu\ =\ 0^{+}} \\[5mm] = &\ \bbx{{1 \over 2}\,\ln^{2}\pars{2} - \gamma\ln\pars{2}} \end{align}