Infinite series containing positive and negative terms 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
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\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}
