How to calculate the value of $\sum\limits_{k=0}^{\infty}\frac{1}{(3k+1)\cdot(3k+2)\cdot(3k+3)}$?

Question

How do I calculate the value of the series $$\sum_{k=0}^{\infty}\frac{1}{(3k+1)\cdot(3k+2)\cdot(3k+3)}= \frac{1}{1\cdot2\cdot3}+\frac{1}{4\cdot5\cdot6}+\frac{1}{7\cdot8\cdot9}+\cdots?$$

Answer 1

By making use of the integral $$\int_{0}^{1} \frac{(1-x)^2}{1-x^3} \, dx = \frac{1}{2} \, \left(\frac{\pi}{\sqrt{3}} - \ln 3 \right)$$ one can take the following path. \begin{align} S &= \sum_{k=0}^{\infty} \frac{1}{(3k+1)(3k+2)(3k+3)} \\ &= \sum_{k=0}^{\infty} \frac{\Gamma(3k+1)}{\Gamma(3k+4)} = \frac{1}{2} \, \sum_{k=0}^{\infty} B(3, 3k+1), \end{align} where $B(n,m)$ is the Beta function, which leads to \begin{align} S &= \frac{1}{2} \, \sum_{k=0}^{\infty} \, \int_{0}^{1} t^{2} \, (1-t)^{3k} \, dt \\ &= \frac{1}{2} \, \int_{0}^{1} \frac{t^{2} \, dt}{1- (1-t)^{3}} \\ &= \frac{1}{2} \, \int_{0}^{1} \frac{(1-x)^{2} \, dx}{1- x^3} \hspace{15mm} x = 1 - t \\ &= \frac{1}{4} \, \left(\frac{\pi}{\sqrt{3}} - \ln 3 \right). \end{align}

Answer 2

The answer is $\frac{\pi\sqrt3}{12}-\frac{\ln3}{4}.$

See the similar problem (problem 2) here: http://www.imc-math.org.uk/imc2010/imc2010-day1-solutions.pdf

Answer 3

You can calculate the partial sum using $$ \frac{1}{(3k+1)(3k+2)(3k+3)}=\frac{1}{2}\frac{1}{3k+1}-\frac{1}{3k+2}+\frac{1}{2}\frac{1}{3k+3} $$

Answer 4

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{\ds{% \sum_{k = 0}^{\infty}{1 \over \pars{3k + 1}\pars{3k + 2}\pars{3k + 3}}}} = {1 \over 27}\sum_{k = 0}^{\infty} {1 \over \pars{k + 1/3}\pars{k + 2/3}\pars{k + 1}} \\[5mm] = & {1 \over 6}\ \underbrace{\sum_{k = 0}^{\infty}\pars{{1 \over k + 1/3} - {1 \over k + 2/3}}} _{\ds{\underbrace{H_{-1/3} - H_{-2/3}} _{Euler\ Reflection\ Formula:\\ \ds{=\ \pi\cot\pars{\pi/3} =\root{3}\pi/3}}}}\ +\ {1 \over 6} \underbrace{\sum_{k = 0}^{\infty}\pars{{1 \over k + 1} - {1 \over k + 2/3}}} _{\ds{H_{-1/3} - H_{0}}}\quad \pars{~H_{z}:\ Harmonic\ Number~} \\[5mm] = &\ {\root{3} \over 18}\,\pi + {1 \over 6}\,H_{-1/3} \end{align}

Moreover, $\ds{H_{-1/3} = H_{2/3} - 3/2\ \pars{~recurrence~}}$.

$\ds{H_{-1/3} = \overbrace{\braces{3\bracks{1 - \ln\pars{3}}/2 + \root{3}\pi/6}}^{\ds{H_{2/3}}}\ -\ 3/2 = -3\ln\pars{3}/2 + \root{3}\pi/6}$. The $\ds{H_{2/3}}$ value is given in a table. Otherwise, it can be evaluated by means of the Gauss Digamma Theorem.

Finally, \begin{align} &\bbox[10px,#ffd]{\ds{% \sum_{k = 0}^{\infty}{1 \over \pars{3k + 1}\pars{3k + 2}\pars{3k + 3}}}} = {\root{3} \over 18}\,\pi + {1 \over 6}\bracks{- {3\ln\pars{3} \over 2} + {\root{3} \over 6}\,\pi} \\[5mm] = &\ \bbx{\root{3}\pi - 3\ln\pars{3} \over 12} \approx 0.1788 \end{align}