Sum of the series
$$\frac {4}{10}+\frac {4\cdot7}{10\cdot20}+ \frac {4\cdot7\cdot10}{10\cdot20\cdot30}+\cdots $$
But i want to solve it using Beta ganmma function.
My Process follows
Let $$S = \frac{1}{10}\bigg[\frac{ 4}{1}+\frac{4\cdot 7}{1\cdot 2}+\frac{4\cdot 7 \cdot 10}{1\cdot 2 \cdot 3}+\cdots \cdots \bigg]$$
Let $\displaystyle a_{n} = \prod^{n}_{k=1}(3k+1)=3^n \prod^{n}_{k=1}\bigg(k+\frac{1}{3}\bigg)$
$\displaystyle =3^n\Gamma\bigg(n+\frac{4}{3}\bigg)\cdot \frac{1}{\Gamma \bigg(\frac{4}{3}\bigg)}$
And Let $\displaystyle b_{n} = \prod^{n}_{k=1}k = \Gamma (n+1)$
So $\displaystyle S =\frac{1}{10} \sum^{\infty}_{n=1}3^n \frac{\Gamma \bigg(n+\frac{4}{3}\bigg)\cdot \Gamma \bigg(\frac{2}{3}\bigg)}{\Gamma \bigg(\frac{2}{3}\bigg)\Gamma \bigg(\frac{4}{3}\bigg)\cdot \Gamma(n+1)}$
using $\displaystyle \Gamma(x)\cdot \Gamma(1-x) = \frac{\pi}{\sin (\pi x)}$ and
$\displaystyle \frac{\Gamma (a) \Gamma(b)}{\Gamma(a+b)} =B(a,b)= \int^{1}_{0}x^{a-1}(1-x)^{b-1}dx$
So $\displaystyle S = \frac{1}{10}\sum^{\infty}_{n=1}3^n\int^{1}_{0}x^{n+\frac{1}{3}}(1-x)^{-\frac{1}{3}}dx$
$\displaystyle =\frac{1}{10}\int^{1}_{0}\frac{3x}{1-3x}\cdot \bigg(\frac{x}{1-x}\bigg)^{\frac{1}{3}}dx$
Now put $\displaystyle \frac{x}{1-x} = t\Rightarrow x = \frac{t}{1+t} = 1-\frac{1}{1+t}$
So $\displaystyle S = \frac{3}{10}\int^{\infty}_{0}\frac{t^{\frac{4}{3}}}{(2t-1)(1+t)^2}dt$
Could some Help me to solve it, Thanks
although it has solved Here
Best Answer
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align} &\bbox[10px,#ffd]{\ds{\sum_{n = 1}^{\infty} {\prod_{k = 1}^{n}\pars{3k + 1} \over \prod_{k = 1}^{n}10k}}} = \sum_{n = 1}^{\infty} {3^{n}\prod_{k = 1}^{n}\pars{k + 1/3} \over 10^{n}n!} = \sum_{n = 1}^{\infty} {\pars{3/10}^{n} \over n!}\,\pars{4 \over 3}^{\large\overline{n}} \\[5mm] = &\ \sum_{n = 1}^{\infty} {\pars{3/10}^{n} \over n!}\,{\Gamma\pars{4/3 + n} \over \Gamma\pars{4/3}} = \sum_{n = 1}^{\infty} \pars{3 \over 10}^{n}\,{\pars{n + 1/3}! \over n!\pars{1/3}!} \\[5mm] = &\ \sum_{n = 1}^{\infty}\pars{3 \over 10}^{n}\,{n + 1/3 \choose n} = \sum_{n = 1}^{\infty}\pars{3 \over 10}^{n} \bracks{{-4/3 \choose n}\pars{-1}^{n}} \\[5mm] = &\ \sum_{n = 1}^{\infty} {-4/3 \choose n}\pars{-\,{3 \over 10}}^{n} = \bracks{1 + \pars{-\,{3 \over 10}}}^{-4/3} - 1 \\[5mm] = &\ \bbx{\pars{10 \over 7}^{4/3} - 1} \approx 0.6089 \end{align}