Serie $\sum _{n=1}^{\infty \:}\ln\left(\frac{n\left(n+2\right)}{\left(n+1\right)^2}\right)$ convergen?

Question

Serie $\sum _{n=2}^{\infty \:}\ln\left(\frac{n\left(n+2\right)}{\left(n+1\right)^2}\right)$ convergen?

Mi idea es aplicar la prueba de Cauchy, pero no sé cómo simplificarla a continuación.

Gracias

Answer 1

Utilizar la identidad $$\prod_{n\geq0}\frac{\left(n+a\right)\left(n+b\right)}{\left(n+c\right)\left(n+d\right)}=\frac{\Gamma\left(c\right)\Gamma\left(d\right)}{\Gamma\left(a\right)\Gamma\left(b\right)},\,a+b=c+d$$ tenemos $$\prod_{n\geq2}\frac{n\left(n+2\right)}{\left(n+1\right)^{2}}=\prod_{n\geq0}\frac{\left(n+2\right)\left(n+4\right)}{\left(n+3\right)^{2}}=\frac{\Gamma^{2}\left(3\right)}{\Gamma\left(2\right)\Gamma\left(4\right)}=\frac{2}{3}.$$ así que $$\sum_{n\geq2}\log\left(\frac{n\left(n+2\right)}{\left(n+1\right)^{2}}\right)=\log\left(\prod_{n\geq2}\frac{n\left(n+2\right)}{\left(n+1\right)^{2}}\right)=\color{red}{\log\left(\frac{2}{3}\right)}.$$

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\sum_{n = 2}^{\infty}\ln\pars{n\bracks{n + 2} \over \bracks{n + 1}^{\,2}} = \sum_{n = 3}^{\infty}\ln\pars{1 - {1 \over n^{2}}} = \sum_{n = 3}^{\infty}\bracks{-2\int_{0}^{1}{t \over n^{2} - t^{2}}\,\dd t} \\[5mm] = &\ -2\int_{0}^{1}t\sum_{n = 3}^{\infty}{1 \over n^{2} - t^{2}}\,\dd t = -\int_{0}^{1}\sum_{n = 3}^{\infty}\pars{{1 \over n - t} - {1 \over n + t}} \,\dd t \\[5mm] = &\ -\int_{0}^{1}\sum_{n = 0}^{\infty} \pars{{1 \over n + 3 - t} - {1 \over n + 3 + t}}\,\dd t \\[5mm] = &\ -\int_{0}^{1}\bracks{\Psi\pars{3 + t} - \Psi\pars{3 - t}}\,\dd t\qquad\qquad \pars{~\Psi:\ Digamma\ Function~} \\[5mm] = &\ \left.\vphantom{\LARGE A}-\ln\pars{\vphantom{\Large A}\Gamma\pars{3 + t}\Gamma\pars{3 - t}} \right\vert_{\ 0}^{\ 1}\qquad\qquad\qquad\pars{~\Gamma:\ Gamma\ Function~} \\[5mm] = &\ -\ln\pars{\Gamma\pars{4}\Gamma\pars{2} \over \Gamma^{2}\pars{3}} = -\ln\pars{6 \times 1 \over 2^{2}} = \bbx{\ds{\ln\pars{2 \over 3}}} \end{align}