An alternative way to find the sum of this series? $\displaystyle \frac{4}{20}$+$\displaystyle \frac{4.7}{20.30}$+$\displaystyle \frac{4.7.10}{20.30.40}$+...
Now  I  have  tried  to  solve  this  in  a  usual  way,  first  find  the  nth  term  $t_n$.
$t_n$=  $\displaystyle \frac{1}{10}$($\displaystyle \frac{1+3}{2}$)  +  $\displaystyle \frac{1}{10^2}$($\displaystyle \frac{1+3}{2}$)($\displaystyle \frac{1+6}{3}$)  +  ...+  $\displaystyle \frac{1}{10^n}$($\displaystyle \frac{1+3}{2}$)($\displaystyle \frac{1+6}{3}$)...($\displaystyle \frac{1+3n}{n+1}$)
=$\displaystyle \frac{1}{10^n}\prod$(1+$\displaystyle \frac{2r}{r+1}$)  ,  $r=1,2,..,n$
=$\displaystyle \prod$($\displaystyle \frac{3}{10}-\displaystyle \frac{1}{5(r+1)}$)
thus,  $t_n=$ (x-$\displaystyle \frac{a}{2}$)(x-$\displaystyle \frac{a}{3}$)...(x-$\displaystyle \frac{a}{n+1}$),  x=$\displaystyle \frac{3}{10}$,  a=$\displaystyle \frac{1}{5}$
Now  to  calculate  $S_n$,  I  have  to  find  the  product  $t_n$,  and  then  take  sum  over  it.  But  this  seems  to  be  a  very  tedious  job.  Is  there  any  elegant  method(may  be  using  the  expansions  of  any  analytic  functions)  to  do  this?
 A: Through Euler's Beta function and the reflection formula for the $\Gamma$ function:
$$\sum_{n\geq 1}\frac{\prod_{k=1}^{n}(3k+1)}{10^n(n+1)!}=\sum_{n\geq 1}\frac{3^n\Gamma\left(n+\frac{4}{3}\right)}{10^n \Gamma(n+2)\Gamma\left(\frac{4}{3}\right)}=\frac{3\sqrt{3}}{2\pi}\sum_{n\geq 1}\left(\tfrac{3}{10}\right)^n B\left(\tfrac{2}{3},n+\tfrac{4}{3}\right) $$
where
$$ \sum_{n\geq 1}\left(\tfrac{3}{10}\right)^n B\left(\tfrac{2}{3},n+\tfrac{4}{3}\right) = \int_{0}^{1}\sum_{n\geq 1}\left(\tfrac{3}{10}\right)^n(1-x)^{-1/3}x^{n+1/3}\,dx=\int_{0}^{1}\frac{3x^{4/3}\,dx}{(1-x)^{1/3}(10-3x)} $$
and the last integral can be computed in a explicit way with a bit of patience. The final outcome is
$$\sum_{n\geq 1}\frac{\prod_{k=1}^{n}(3k+1)}{10^n(n+1)!}=\color{red}{10\sqrt[3]{\frac{10}{7}}-11} $$
which can also be proved by invoking Lagrange's inversion theorem or the extended binomial theorem.
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]{\sum_{n = 1}^{\infty}{\prod_{k = 1}^{n}\pars{3k + 1} \over 10^{n}\pars{n + 1}!}} =
\sum_{n = 2}^{\infty}{3^{n - 1}
\prod_{k = 1}^{n - 1}\pars{k + 1/3} \over 10^{n - 1}\,n!}
\\[5mm] = &\
{10 \over 3}\sum_{n = 2}^{\infty}\pars{3 \over 10}^{n}\,
{\Gamma\pars{4/3 + \bracks{n - 1}}/\Gamma\pars{4/3} \over n!}
\\[5mm] = &\
{10 \over 3}\,{\pars{-2/3}! \over \Gamma\pars{4/3}}
\sum_{n = 2}^{\infty}\pars{3 \over 10}^{n}\,
{\pars{n - 2/3}! \over n!\pars{-2/3}!}
\\[5mm] = &\
{10 \over 3}\,{\Gamma\pars{1/3} \over \pars{1/3}\Gamma\pars{1/3}}
\sum_{n = 2}^{\infty}\pars{3 \over 10}^{n}\,{n - 2/3 \choose n}
\\[5mm] = &\
10\sum_{n = 2}^{\infty}\pars{3 \over 10}^{n}
\bracks{{-1/3 \choose n}\pars{-1}^{n}}
\\[5mm] = &\
10\bracks{%
\sum_{n = 0}^{\infty}{-1/3 \choose n}\pars{-\,{3 \over 10}}^{n}
-\ \overbrace{-1/3 \choose 0}^{\ds{= 1}}\ -\
\overbrace{-1/3 \choose 1}^{\ds{= -\,{1 \over 3}}}\
\pars{-\,{3 \over 10}}}
\\[5mm] = &\
10\braces{\bracks{1 + \pars{-\,{3 \over 10}}}^{-1/3} - 1 - {1 \over 10}}
=
\bbx{10\pars{10 \over 7}^{1/3} - 11}
\\[5mm] \approx &\ 0.2625
\end{align}
