Cómo mostrar que $\sum_{n=1}^{\infty} \frac{1}{(2n+1)(2n+2)(2n+3)}=\ln(2)-1/2$?

Question

Cómo puedo demostrar que

$$ \sum_{n=1}^{\infty} \frac{1}{(2n+1)(2n+2)(2n+3)}=\ln(2)-1/2 $$

Y $$ \sum_{n=1}^{\infty} \frac{1}{(4n+1)(4n+2)(4n+3)(4n+4)}=\frac{1}{4}\left(\ln(2) - \frac{\pi}{6}\right). $$

Gracias de antemano.

Answer 1

Un enfoque alternativo; calcular

$$f(x)=\sum_{n=0}^\infty \frac{x^{2n+3}}{(2n+3)(2n+2)(2n+1)}$$

Este tiene la propiedad de que $f'''(x)= \sum_{n=0}^\infty x^{2n} = \frac{1}{1-x^2}$. Integrar la vez (parcial fracciones) para obtener los logaritmos. Integrar dos veces más (por partes) y evaluar en 1 para obtener la respuesta $f(1)$.

Answer 2

Esto es similar, pero sin parcial de las fracciones.

$$(1)$$

$$ \begin{aligned} \sum_{n\geq 0}\frac{1}{(2n+1)(2n+2)(2n+3)} & =\sum_{n\geq 0} \frac{\Gamma(2n+1)}{\Gamma(2n+4)} \\& =\sum_{n\geq 0}\frac{1}{2}\mathrm{B}(2n+1, \,3) \\& = \frac{1}{2}\sum_{n\geq 0}\int_{0}^{1}x^{2n}(1-x)^2\; dx\\& = \frac{1}{2}\int_{0}^{1} \frac{1-x}{1+x} \;{dx} \\& = \ln 2-\frac{1}{2} \end{aligned}$$

$$(2)$$

$$ \begin{aligned} \sum_{n\geq 0}\frac{1}{(4n+1)(4n+2)(4n+3)(4n+4)} & =\sum_{n\geq 0} \frac{\Gamma(4n+1)}{\Gamma(4n+5)} \\& =\sum_{n\geq 0}\frac{1}{6}\mathrm{B}(4n+1, \,4) \\& = \frac{1}{6}\sum_{n\geq 0}\int_{0}^{1}x^{4n}(1-x)^3\; dx\\& = \frac{1}{6}\int_{0}^{1} \frac{(1-x)^2}{(1+x^2)(1+x)} \; dx \\& = \frac{\ln 2}{4}-\frac{\pi}{24} \end{aligned}$$

Answer 3

$$ \begin{align} &\sum_{n=0}^\infty\frac1{(2n+1)(2n+2)(2n+3)}\\ &=\sum_{n=0}^\infty\left(\frac{1/2}{2n+1}+\frac{-1}{2n+2}+\frac{1/2}{2n+3}\right)\\ &=-\frac{\frac12}{1}+\frac{\frac12}{1}\tag{add %#%#%}\\ &\hphantom{=-\frac{\frac12}{1}}+\frac{\frac12}{1}-\frac12+\frac{\frac12}{3}\tag{%#%#%}\\ &\hphantom{=-\frac{\frac12}{1}+\frac{\frac12}{1}-\frac12}+\frac{\frac12}{3}-\frac14+\frac{\frac12}{5}\tag{%#%#%}\\ &\hphantom{=-\frac{\frac12}{1}+\frac{\frac12}{1}-\frac12+\frac{\frac12}{3}-\frac14}+\frac{\frac12}{5}-\frac16+\dots\tag{%#%#%}\\ &=\color{#C00000}{-\frac12}+\color{#00A000}{1-\frac12\,+\frac13-\frac14+\frac15-\frac16+\dots}\\ &=\color{#00A000}{\log(2)}\color{#C00000}{-\frac12} \end{align} $$

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$$ \begin{align} &\sum_{n=0}^\infty\frac1{(4n+1)(4n+2)(4n+3)(4n+4)}\\ &=\sum_{n=0}^\infty\left(\frac{1/6}{4n+1}+\frac{-1/2}{4n+2}+\frac{1/2}{4n+3}+\frac{-1/6}{4n+4}\right)\\ &=\sum_{n=0}^\infty\left(\frac{-1/6}{4n+1}+\frac{1/6}{4n+3}\right)+\left(\frac{1/3}{4n+1}+\frac{1/3}{4n+3}\right)\\ &\hphantom{\sum_{n=0}^\infty}+\left(\frac{-1/6}{4n+2}+\frac{1/6}{4n+4}\right)+\left(\frac{-1/3}{4n+2}+\frac{-1/3}{4n+4}\right)\\ &=\color{#00A000}{\sum_{n=0}^\infty\left(\frac{-1/6}{4n+1}+\frac{1/6}{4n+3}\right)}+\color{#0000FF}{\sum_{n=0}^\infty\left(\frac{-1/6}{4n+2}+\frac{1/6}{4n+4}\right)}\\ &+\color{#C00000}{\sum_{n=0}^\infty\left(\frac{1/3}{4n+1}+\frac{1/3}{4n+3}\right)+\left(\frac{-1/3}{4n+2}+\frac{-1/3}{4n+4}\right)}\\ &=\color{#00A000}{-\frac\pi{24}}\color{#0000FF}{-\frac{\log(2)}{12}}\\ &\color{#C00000}{+\frac{\log(2)}{3}}\\ &=\frac{\log(2)}{4}-\frac\pi{24} \end{align} $$ $0$, $n=0$, e $n=1$.

Answer 4

\begin{align} &\sum_{n = 0}^{\infty} {1 \over \left(2n + 1\right)\left(2n + 2\right)\left(2n + 3\right)} = {1 \over 4}\sum_{n = 0}^{\infty} \left({1 \over n + 1/2} - {2 \over n + 1} + {1 \over n + 3/2}\right) \\[3mm]&= {1 \over 8}\left\lbrack \sum_{n = 0}^{\infty}{1 \over \left(n + 1/2\right)\left(n + 1\right)} - \sum_{n = 0}^{\infty}{1 \over \left(n + 1\right)\left(n + 3/2\right)} \right\rbrack \\[3mm]&= {1 \over 8}\left\lbrack {\Psi\left(1/2\right) - \Psi\left(1\right) \over 1/2 - 1} - {\Psi\left(1\right) - \Psi\left(3/2\right) \over 1 - 3/2} \right\rbrack = {1 \over 4}\left\lbrack - \Psi\left(1 \over 2\right) + 2\Psi\left(1\right) - \Psi\left(3 \over 2\right) \right\rbrack \\[3mm]&= {1 \over 4}\left\lbrace -\Psi\left(1 \over 2\right) + 2\Psi\left(1\right) - \left\lbrack 2 + \Psi\left(1 \over 2\right)\right\rbrack \right\rbrace = {1 \over 2}\left\lbrack -\Psi\left(1 \over 2\right) + \Psi\left(1\right) - 1 \right\rbrack \\[3mm]&= {1 \over 2}\left\lbrace -\left\lbrack-\gamma - 2\ln\left(2\right)\right\rbrack + \left(-\gamma\right) - 1 \right\rbrace = \ln\left(2\right) - {1 \over 2} \\[1cm]& \end{align}

\begin{align} \sum_{n = 0}^{\infty}{1 \over \left(2n + 1\right)\left(2n + 2\right)\left(2n + 3\right)} &= \ln\left(2\right) - {1 \over 2} \\[3mm] \color{#0000FF}{\large\sum_{n = 1}^{\infty}{1 \over \left(2n + 1\right)\left(2n + 2\right)\left(2n + 3\right)}} &= \left\lbrack\ln\left(2\right) - {1 \over 2}\right\rbrack - {1 \over 6} = \color{#0000ff}{\large \ln\left(2\right) - {2 \over 3}} \\[1cm]& \end{align}

$\Psi\left(z\right)$ es la función Digamma y $\gamma = 0.57721566490153286060651209008240243104215933593992\ldots$ es el de Euler-Mascheroni constante.