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A series is given as follows

$$\frac{1}{6} + \frac{5}{6\cdot 12} + \frac{5\cdot 8}{6\cdot 12\cdot 18} + \frac{5\cdot 8\cdot 11}{6\cdot 12\cdot 18\cdot 24}+\ldots$$

Can you give me hints to get started finding its value? Thanks.

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  • $\begingroup$ 6.12.18, are you sure? $\endgroup$ – Jorge Fernández Hidalgo Jan 5 '15 at 5:20
  • $\begingroup$ Can you please edit your thread to use LaTeX? It will make things much clearer. $\endgroup$ – ml0105 Jan 5 '15 at 5:20
  • $\begingroup$ It would be nice if you would properly format your question. $\endgroup$ – Gahawar Jan 5 '15 at 5:20
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    $\begingroup$ I think the intent is $\sum_{n=1}^\infty \frac{\prod_{i=0}^{n-2} 5+3i}{\prod_{i=1}^n 6i}$, where my convention is that the empty product $\prod_{i=0}^{-1} a_i$ is $1$. $\endgroup$ – Ian Jan 5 '15 at 5:25
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    $\begingroup$ You can recognize this series as a Taylor series for $\frac12(x+1)^{-2/3}-\frac12$ evaluated at $x=-{\frac12}$. The general way to handle this kind of series is to (1) recognize that the denominator has $n!$ that can be factored out, which is part of the taylor formula, and (2) recognize the numerator's factors increase by $3$. So with appropriate manipulation, you can find a series for a function of the form $(1+x)^{k/3}$ and an $x$-value that almost gives you your series. (There is also a scaling factor and an accounting of the constant term to work out.) $\endgroup$ – alex.jordan Jan 5 '15 at 6:28
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Consider the binomial expansion of $(1+x)^n=1+nx+\frac{n(n-1)}{2}x^2+...$ Multiplying the series term by term by $2$

On comparing, $nx=\frac{1}{3}$ and $\frac{n(n-1)}{2}x^2=\frac{5}{3.12}$

Solving for $n,x$ we get $n=\frac{-2}{3},x=\frac{-1}{2}$

So the sum becomes $\{(\frac{1}{2})^{\frac{-2}{3}}-1\}=2^\frac{2}{3}-1=4^\frac{1}{3}-1$

since $2$ has been multiplied before divide the result by $2$

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  • $\begingroup$ thanks for answer , but answer is what as stated by oleg567 in comment $\endgroup$ – Sophie Clad Jan 5 '15 at 5:51
  • $\begingroup$ @SophieClad check if this suits you $\endgroup$ – Learnmore Jan 5 '15 at 6:26
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$$\begin{align} \sum_{n=1}^{\infty}\frac12\prod_{i=1}^n\frac{3i-1}{6i} &=\frac12\sum_{n=1}^{\infty}\frac1{2^nn!}\prod_{i=1}^n\frac{3i-1}{3}\\ &=\frac12\sum_{n=1}^{\infty}\frac1{(-2)^nn!}\prod_{i=1}^n\frac{1-3i}{3}\\ &=\frac12\sum_{n=1}^{\infty}\frac1{(-2)^nn!}\prod_{i=1}^n\left(\frac13-i\right)\\ &=\left[\frac12\sum_{n=1}^{\infty}\frac{x^n}{n!}\prod_{i=1}^n\left(\frac13-i\right)\right]_{x=-1/2}\\ \end{align}$$

Now note that for $n\geq1$, $\prod_{i=1}^n\left(\frac13-i\right)=f^{(n)}(0)$, where $f(x)=(x+1)^{-2/3}$. We can manipulate the expression a little further to see it as a Taylor series for $f$.

$$\begin{align} \sum_{n=1}^{\infty}\frac12\prod_{i=1}^n\frac{3i-1}{6i} &=\left[\frac12\sum_{n=0}^{\infty}\frac{x^n}{n!}\prod_{i=1}^n\left(\frac13-i\right)-\frac12\right]_{x=-1/2}\\ &=\left[\frac12\sum_{n=0}^{\infty}\frac{x^n}{n!}f^{(n)}(0)\right]_{x=-1/2}-\frac12\\ &=\frac12f(-1/2)-\frac12\\ &=\frac12(1/2)^{-2/3}-\frac12\\ &=\sqrt[3]{1/2}-\frac12\\ \end{align}$$

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For any $n\in\mathbb{N}^+$ we have $$ \prod_{k=1}^{n}(3k+2) = \frac{3^n\,\Gamma\left(n+\frac{5}{3}\right)}{\Gamma\left(\frac{5}{3}\right)}\tag{1}$$ and the value of the RHS at $n=0$ equals $1$. It follows that the whole series can be represented as

$$ S=\sum_{n\geq 0}\frac{1}{6^{n+1}(n+1)!}\cdot\frac{3^n\,\Gamma\left(n+\frac{5}{3}\right)}{\Gamma\left(\frac{5}{3}\right)}=\sum_{n\geq 0}\frac{1}{6\cdot 2^n}\cdot\frac{\Gamma\left(n+\frac{5}{3}\right)}{\Gamma\left(\frac{5}{3}\right)\Gamma(n+2)}\tag{2} $$ or, by using Euler's beta function and the reflection formula for the $\Gamma$ function, as $$ \sum_{n\geq 0}\frac{1}{6\,\Gamma\left(\frac{1}{3}\right)\Gamma\left(\frac{5}{3}\right)}\cdot\frac{B\left(n+\frac{5}{3},\frac{1}{3}\right)}{2^n}=\frac{\sqrt{3}}{8\pi}\sum_{n\geq 0}\int_{0}^{1}\frac{x^{n+2/3}(1-x)^{-2/3}}{2^n}\,dx \tag{3}$$ from which: $$ S = \frac{\sqrt{3}}{4\pi}\int_{0}^{1}\frac{x^{2/3}}{(1-x)^{2/3}(2-x)}\,dx \stackrel{\frac{x}{1-x}\mapsto z}{=}\frac{\sqrt{3}}{4\pi}\int_{0}^{+\infty}\frac{z^{2/3}}{(1+z)(2+z)}\tag{4}$$ and: $$ S = \frac{\sqrt{3}}{4\pi}\left(-\int_{0}^{+\infty}\frac{z^{2/3}}{z(1+z)}\,dz+2\int_{0}^{+\infty}\frac{z^{2/3}}{z(2+z)}\,dz\right).\tag{5} $$ Euler's beta function then leads to: $$ S = \color{blue}{\frac{1}{\sqrt[3]{2}}-\frac{1}{2}} \approx 0.2937.\tag{6}$$

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  • $\begingroup$ please explain me how can we write $\displaystyle \prod_{k=1}^{n}(3k+2) = \frac{3^n\,\Gamma\left(n+\frac{5}{3}\right)}{\Gamma\left(\frac{5}{3}\right)}$. $\endgroup$ – DXT Sep 15 '18 at 7:21
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    $\begingroup$ @DurgeshTiwari: by just exploiting the main functional identity for the $\Gamma$ function: $$\Gamma(n+5/3)=(n+2/3)\Gamma(n+2/3)=(n+2/3)(n-1/3)\Gamma(n-1/3)=\ldots$$ $\endgroup$ – Jack D'Aurizio Sep 15 '18 at 16:33
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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} &{1 \over 6} + {5 \over 6 \cdot 12} + {5 \cdot 8 \over 6 \cdot 12 \cdot 18} + {5 \cdot 8 \cdot 11 \over 6\cdot 12\cdot 18\cdot 24} + \cdots = {1 \over 6} + \sum_{n = 0}^{\infty}{\prod_{k = 0}^{n}\pars{3k + 5} \over \prod_{k = 0}^{n + 1}\pars{6k + 6}} \\[5mm] = &\ {1 \over 6} + \sum_{n = 0}^{\infty}{3^{n + 1}\prod_{k = 0}^{n}\pars{k + 5/3} \over 6^{n + 2}\prod_{k = 0}^{n + 1}\pars{k + 1}} = {1 \over 6} + {1 \over 6}\sum_{n = 0}^{\infty}{1 \over 2^{n + 1}}\,{\pars{5/3}^{\overline{n + 1}} \over \pars{n + 2}!} \\[5mm] = &\ {1 \over 6} + {1 \over 6}\sum_{n = 0}^{\infty}{1 \over 2^{n + 1}}\, {\Gamma\pars{n + 8/3}/\Gamma\pars{5/3} \over \pars{n + 2}!} = {1 \over 6} + {1 \over 6}\sum_{n = 0}^{\infty}{1 \over 2^{n + 1}}\, {\pars{n + 5/3}! \over \pars{2/3}!\pars{n + 2}!} \\[5mm] = &\ {1 \over 6} + {1 \over 6}\sum_{n = 0}^{\infty}{1 \over 2^{n + 1}} {n + 8/3 \choose n + 2}{1 \over n + 8/3} = {1 \over 6} + {1 \over 6}\sum_{n = 0}^{\infty}{1 \over 2^{n + 1}} {-5/3 \choose n + 2}\pars{-1}^{n + 2}\,{1 \over n + 8/3} \\[5mm] = &\ {1 \over 6} + {1 \over 6}\sum_{n = 2}^{\infty}{1 \over 2^{n - 1}} {-5/3 \choose n}\pars{-1}^{n}\,{1 \over n + 2/3} = {1 \over 6} + {1 \over 6}\sum_{n = 2}^{\infty}{\pars{-1}^{n} \over 2^{n - 1}} {-5/3 \choose n}\int_{0}^{1}x^{n - 1/3}\,\dd x \\[5mm] = &\ {1 \over 6} + {1 \over 3}\int_{0}^{1} \sum_{n = 2}^{\infty}{-5/3 \choose n}\pars{-\,{x \over 2}}^{n} \,{\dd x \over x^{1/3}} = {1 \over 6} + {1 \over 3}\ \underbrace{\int_{0}^{1} \bracks{\pars{1 - {x \over 2}}^{-5/3} - 1 - {5 \over 3}\,{x \over 2}} \,{\dd x \over x^{1/3}}}_{\ds{-2 + {3 \over 2^{1/3}}}} \\[5mm] = &\ \bbx{{1 \over 2^{1/3}} - {1 \over 2}} \approx 0.2937 \end{align}

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