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?

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.
