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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 '18 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 '18 at 20:23
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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 '18 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 '18 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 '18 at 18:29
  • $\begingroup$ Ah, I see. Thanks for the clarification. $\endgroup$ – Jack Moody Feb 10 '18 at 18:30
  • $\begingroup$ you can't use this directly: all individual terms are divergent. $\endgroup$ – dezdichado Feb 10 '18 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 '18 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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