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Calculation of sum of

$\displaystyle 1-\frac{1}{5}+\frac{1\cdot 4}{5\cdot 10}-\frac{1\cdot 4 \cdot 7}{5\cdot 10 \cdot 15}+\cdots \cdots $

The series is not in arithmetic series or in arithmetic geometric series

I did not understand how to solve such type of problem

help me to solve it plaese

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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]{1 - \sum_{n = 0}^{\infty}\pars{-1}^{n}\, {\prod_{k = 0}^{n}\pars{3k + 1} \over \prod_{k = 0}^{n}\pars{5k + 5}}} \\[5mm] = &\ 1 - \sum_{n = 0}^{\infty}\pars{-1}^{n}\, {3^{n + 1}\prod_{k = 0}^{n}\pars{k + 1/3} \over 5^{n + 1}\prod_{k = 0}^{n}\pars{k + 1}} \\[5mm] = &\ 1 + \sum_{n = 0}^{\infty}\pars{-\,{3 \over 5}}^{n + 1}\, {\pars{1/3}^{\overline{n + 1}} \over \pars{n + 1}!} \\[5mm] = &\ 1 + \sum_{n = 0}^{\infty} {\Gamma\pars{1/3 + n + 1}/\Gamma\pars{1/3} \over \pars{n + 1}!}\, \pars{-\,{3 \over 5}}^{n + 1} \\[5mm] = &\ 1 + \sum_{n = 0}^{\infty} {\pars{n + 1/3}! \over \pars{n + 1}!\pars{-2/3}!}\, \pars{-\,{3 \over 5}}^{n + 1} = 1 + \sum_{n = 0}^{\infty}{n + 1/3 \choose n + 1} \pars{-\,{3 \over 5}}^{n + 1} \\[5mm] = &\ 1 + \sum_{n = 0}^{\infty} \bracks{{-1/3 \choose n + 1}\pars{-1}^{n + 1}} \pars{-\,{3 \over 5}}^{n + 1} \\[5mm] = &\ \sum_{n = 0}^{\infty} {-1/3 \choose n}\pars{3 \over 5}^{n} = \pars{1 + {3 \over 5}}^{-1/3} = \bbx{5^{1/3} \over 2} \approx 0.8550 \end{align}

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  • $\begingroup$ cool the passage through binomial (+1) $\endgroup$ – G Cab Dec 5 '18 at 22:13
  • $\begingroup$ @GCab Thanks. Binomials are always quite fine. $\endgroup$ – Felix Marin Dec 5 '18 at 22:31

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