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

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    $\begingroup$ The answer is $\frac{\pi\sqrt3}{12}-\frac{\ln3}{4}.$ $\endgroup$
    – Michael Rozenberg
    Feb 10, 2018 at 18:38
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    $\begingroup$ I have changed the formatting of the title so as to make it take up less vertical space -- this is a policy to ensure that the scarce space on the main page is distributed evenly over the questions. See here for more information. Please take this into consideration for future questions. Thanks in advance. $\endgroup$
    – GNUSupporter 8964民主女神 地下教會
    May 2, 2018 at 20:23

4 Answers 4

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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}

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  • $\begingroup$ How do we justify exchanging the limit of the series and the integral here? AFAIK, the convergence of the series isn't uniform on [0,1]. $\endgroup$
    – Anu
    Feb 10, 2018 at 20:07
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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

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

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  • $\begingroup$ Where does this come from? $\endgroup$
    – Jack Moody
    Feb 10, 2018 at 18:27
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    $\begingroup$ Partial decomposition. For example the coef before $1/(3k+1)$ is $$a=\frac{1}{(-1+2)(-1+3)}=\frac{1}{2}$$. Hope i did not make mistakes for the rest. $\endgroup$
    – Atmos
    Feb 10, 2018 at 18:29
  • $\begingroup$ Ah, I see. Thanks for the clarification. $\endgroup$
    – Jack Moody
    Feb 10, 2018 at 18:30
  • $\begingroup$ you can't use this directly: all individual terms are divergent. $\endgroup$
    – dezdichado
    Feb 10, 2018 at 18:56
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    $\begingroup$ @Atmos and? There is not any cancellation and the partial sum does not have any closed form. I would be surprised if you can calculate it, just using the decomposition. $\endgroup$
    – dezdichado
    Feb 10, 2018 at 19:05
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}

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