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Evaluate : $$1+\frac{2}{6}+\frac{2\cdot 5}{6\cdot 12}+\frac{2\cdot 5\cdot 8}{6\cdot12\cdot 18}+\ldots$$

thought: I have tried to manipulate the series so that the sum upto n terms , can be simplified. But I can't figure out any profitable result.

Jack D'Aurizio
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Akash Banerjee
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    What about computing the Taylor series of $$ f(x)=(1-x)^{-2/3}$$ ? – Jack D'Aurizio Jun 30 '17 at 19:16
  • ok I am trying it out – Akash Banerjee Jun 30 '17 at 19:17
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    Very similar to https://math.stackexchange.com/questions/2327157/find-sum-of-series-frac16-frac56-cdot12-frac5-cdot86-cdot12-cdo/2327192#2327192 – Donald Splutterwit Jun 30 '17 at 19:21
  • Awesome!!! it worked...but how u figured that out ?@ Jack D'Aurizio – Akash Banerjee Jun 30 '17 at 19:22
  • One way to approach: For $\alpha\in\mathbb{R}$ and $n\in\mathbb{N}$, we define the generalized binomial coefficient as the following: $$\binom{\alpha}{n}=\frac{\alpha(\alpha-1)\cdots(\alpha-n+1)}{n!},$$ and $\binom{\alpha}{0}=1$. Moreover, one has the following binomial theorem: $$(1+t)^\alpha=\sum_{n=0}^\infty\binom{\alpha}{n}t^n.$$ – Frank Lu Jun 30 '17 at 19:32
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    @AkashBanerjee: this is a famous problem, to be honest I already knew the most efficient approach. Anyway, a brute-force solution has been outlined below. – Jack D'Aurizio Jun 30 '17 at 19:40

2 Answers2

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It is easier to recognize a Taylor series for some function of the form $(1\pm z)^{\alpha}$, but there also is an interesting approach through Euler's Beta function.

We have $$ \frac{1}{N!6^N}\prod_{n=1}^{N}(3n-1) = \frac{\Gamma\left(N+\frac{2}{3}\right)}{2^N \Gamma(N+1)\,\Gamma\left(\frac{2}{3}\right)}=\frac{\sqrt{3}}{\pi 2^{N+1}}B\left(N+\frac{2}{3},\frac{1}{3}\right) \tag{1}$$ hence $$\begin{eqnarray*} \sum_{N\geq 1}\frac{1}{N!6^N}\prod_{n=1}^{N}(3n-1)&=&\frac{\sqrt{3}}{2\pi}\int_{0}^{1}\sum_{N\geq 1}\frac{(1-x)^{-2/3}x^{N-1/3}}{2^N}\,dx\\&=&\frac{\sqrt{3}}{2\pi}\int_{0}^{1}\frac{x^{2/3}}{(1-x)^{2/3}(2-x)}\,dx\\\left(\frac{x}{1-x}\mapsto w\right)\quad&=&\frac{\sqrt{3}}{2\pi}\int_{0}^{+\infty}\frac{w^{2/3}}{(1+w)(2+w)}\,dw\tag{2} \end{eqnarray*}$$ where the last integral can be easily computed by setting $w=t^3$ and applying partial fraction decomposition. The final outcome is $$\sum_{N\geq 1}\frac{1}{N!6^N}\prod_{n=1}^{N}(3n-1) = \color{red}{2^{2/3}-1} \tag{3}$$ so the given series converges to $\color{red}{\sqrt[3]{4}}$.

Jack D'Aurizio
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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 + {2 \over 6} + {2 \cdot 5 \over 6 \cdot 12} + {2 \cdot 5 \cdot 8 \over 6 \cdot 12 \cdot 18} +\cdots = 1 + \sum_{n = 0}^{\infty}{\prod_{k = 0}^{n}\pars{3k + 2} \over \prod_{k = 0}^{n}\pars{6k + 6}} \\[5mm] & = 1 + \sum_{n = 0}^{\infty}{1 \over 2^{n + 1}\pars{n + 1}!}\ \overbrace{\prod_{k = 0}^{n}\pars{k + {2 \over 3}}} ^{\substack{\ds{\pars{2/3}^{\overline{n + 1}} =}\\[1.5mm] \ds{\Gamma\pars{5/3 + n}/\Gamma\pars{2/3}}}} = 1 + \sum_{n = 0}^{\infty}{1 \over 2^{n + 1}} {\pars{n + 2/3}! \over \pars{n + 1}!\pars{-1/3}!} \\[5mm] & = 1 + \sum_{n = 0}^{\infty}{1 \over 2^{n + 1}}{n + 2/3 \choose n + 1} = 1 + \sum_{n = 0}^{\infty}{1 \over 2^{n + 1}}{-2/3 \choose n + 1}\pars{-1}^{n + 1} = 1 + \sum_{n = 1}^{\infty}{-2/3 \choose n}\pars{-\,{1 \over 2}}^{n} \\[5mm] & = 1 + \braces{\bracks{1 + \pars{-\,{1 \over 2}}}^{-2/3} - {-2/3 \choose 0}} = \bbx{2^{2/3}} \approx 1.5874 \end{align}

Felix Marin
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