La evaluación de $\sum\limits_{k=1}^{\infty}\frac1{(3k+1)(3k+2)}$

Question

¿Cuál es el valor de $\displaystyle\sum_{k=1}^{\infty}\frac1{(3k+1)(3k+2)}$?

Answer 1

Me gusta solución con Gamma/función Beta:

\begin{aligned} \sum_{k=1}^{+\infty}\frac{1}{(3k+1)(3k+2)} & =\sum_{k=1}^{+\infty} \frac{\Gamma(3k+1)}{\Gamma(3k+3)} \\& =\sum_{k=1}^{+\infty}\operatorname{B}(3k+1, 2) \\& = \sum_{k=1}^{+\infty}\int_{0}^{1}x^{3k}(1-x)\, dx\\&= \int_{0}^{1} \sum_{k=1}^{+\infty}x^{3k}(1-x)\, dx\\& = \int_{0}^{1} \frac{x^3}{1+x+x^2} \ dx \\& = \frac{\sqrt{3}\pi}{9}-\frac{1}{2} \end{aligned}

Answer 2

Vamos a observar la relación con el $\;\sin(2\pi j/3)$ y volver a escribir esto en un modo elemental :

\begin{align} \sum_{k=1}^{\infty}\frac1{(3k+1)(3k+2)}&= \sum_{k=1}^{\infty}\frac 0{3k+0}+\frac 1{3k+1}-\frac 1{3k+2}\\ &= \frac 2{\sqrt{3}}\Im\left(\sum_{j=1}^\infty\frac {e^{2\pi\,i\,j/3}}{j}\right)-\frac 1{1\cdot 2}\\ &= -\frac 2{\sqrt{3}}\Im\,\ln\left(1-e^{2\pi\,i/3}\right)-\frac 12\\ &= -\frac 2{\sqrt{3}}\Im\,\ln\left(\frac 32-\frac{\sqrt{3}}2i\right)-\frac 12\\ &= -\frac 2{\sqrt{3}}\Im\,\ln\left(\sqrt{3}\;e^{-\pi\,i/6}\right)-\frac 12\\ &= -\frac 2{\sqrt{3}}\left(-\frac{\pi}6\right)-\frac 12\\ &= \frac {\sqrt{3}\,\pi}{9}-\frac 12\\ \end{align}

Answer 3

\begin{align} &\sum_{k = 0}^{\infty}{1 \over \left(3k + 1\right)\left(3k + 2\right)} = {1 \over 9}\sum_{k = 0}^{\infty}{1 \over \left(k + 1/3\right)\left(k + 2/3\right)} = {1 \over 9}\,{\Psi\left(1/3\right) - \Psi\left(2/3\right) \over 1/3 - 2/3} \\[3mm]&= {{1 \over 3}\left\lbrack \Psi\left(2 \over 3\right) - \Psi\left(1 \over 3\right)\right\rbrack} = {1 \over 3}\,\pi\ {\rm cotan}\left(\pi \over 3\right) = {1 \over 3}\,\pi\ {\sqrt{3} \over 3} \\[5mm]& \end{align} $\Psi\left(z\right)$ es la función Digamma. En el último paso yo uso $\Psi\left(z\right) - \Psi\left(1 - z\right) = -\pi\,{\rm cotan}\left(\pi\,z\right)$ con $z = 1/3$.

$$ \begin{array}{|c|}\hline\\ {\large\quad\sum_{k = 1}^{\infty}{1 \over \left(3k + 1\right)\left(3k + 2\right)} = \sum_{k = 0}^{\infty}{1 \over \left(3k + 1\right)\left(3k + 2\right)} - {1 \over 2} = {\sqrt{3} \over 9}\,\pi - {1 \over 2}\quad} \\ \\ \hline \end{array} $$

Answer 4

El uso de $(7)$ a partir de esta respuesta, $$ \begin{align} \sum_{k=1}^\infty\frac1{(3k+1)(3k+2)} &=\lim_{n\to\infty}\sum_{k=1}^n\left(\frac1{3k+1}-\frac1{3k+2}\right)\\ &=\lim_{n\to\infty}\left(\sum_{k=1}^n\frac1{3k+1}+\sum_{k=-n-1}^{-2}\frac1{3k+1}\right)\\ &=-\frac12+\lim_{n\to\infty}\sum_{k=-n}^n\frac1{3k+1}\\ &=-\frac12+\frac13\sum_{k=-\infty}^\infty\frac1{k+\frac13}\\ &=-\frac12+\frac\pi3\cot\left(\frac\pi3\right)\\ &=\frac\pi{3\sqrt3}-\frac12 \end{align} $$