What is the value of $\displaystyle\sum_{k=1}^{\infty}\frac1{(3k+1)(3k+2)}$?
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1/4-1/5+1/7-1/8+1/10-1/11+... doesn't telescope... – Christmas Bunny Aug 26 '13 at 07:30
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3I think you need to break it down as $$\sum\limits_{j = 1}^\infty {\left( {\frac{1}{{6j - 2}} - \frac{1}{{6j - 1}} + \frac{1}{{6j + 1}} - \frac{1}{{6j + 2}}} \right)} $$ and do something smart. I have seen robjohn do this before. Let me do a search. – Pedro Aug 26 '13 at 07:56
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Here – Pedro Aug 26 '13 at 07:59
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1Please do not use titles consisting only of math expressions; these are discouraged for technical reasons -- see meta. – Lord_Farin Aug 26 '13 at 11:38
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This sum was also evaluated at this MSE link. – Marko Riedel May 20 '14 at 01:38
4 Answers
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I like solution with Gamma/Beta function:
\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}
Cortizol
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Let's observe the relation with $\;\sin(2\pi j/3)$ and rewrite this in an elementary way :
\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}
Raymond Manzoni
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\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} $\Psi\left(z\right)$ is the Digamma function. In the last step I use $\Psi\left(z\right) - \Psi\left(1 - z\right) = -\pi\,{\rm cotan}\left(\pi\,z\right)$ with $z = 1/3$.
$$ \begin{array}{|c|}\hline\\ {\large\quad\sum_{k = 1}^{\infty}{1 \over \left(3k + 1\right)\left(3k + 2\right)} = \sum_{k = 0}^{\infty}{1 \over \left(3k + 1\right)\left(3k + 2\right)} - {1 \over 2} = {\sqrt{3} \over 9}\,\pi - {1 \over 2}\quad} \\ \\ \hline \end{array} $$
Felix Marin
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The value is $\frac{\sqrt{3}}{9} \pi - \frac{1}{2} = 0.104599788078072616864692752548\dots$ – gammatester Aug 26 '13 at 08:31
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Using $(7)$ from this answer, $$ \begin{align} \sum_{k=1}^\infty\frac1{(3k+1)(3k+2)} &=\lim_{n\to\infty}\sum_{k=1}^n\left(\frac1{3k+1}-\frac1{3k+2}\right)\\ &=\lim_{n\to\infty}\left(\sum_{k=1}^n\frac1{3k+1}+\sum_{k=-n-1}^{-2}\frac1{3k+1}\right)\\ &=-\frac12+\lim_{n\to\infty}\sum_{k=-n}^n\frac1{3k+1}\\ &=-\frac12+\frac13\sum_{k=-\infty}^\infty\frac1{k+\frac13}\\ &=-\frac12+\frac\pi3\cot\left(\frac\pi3\right)\\ &=\frac\pi{3\sqrt3}-\frac12 \end{align} $$
