Dtyjlui

$displaystyle frac{4}{20}$+$displaystyle frac{4.7}{20.30}$+$displaystyle frac{4.7.10}{20.30.40}$+...

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:
$$sum_{ngeq 1}frac{prod_{k=1}^{n}(3k+1)}{10^n(n+1)!}=sum_{ngeq 1}frac{3^nGammaleft(n+frac{4}{3}right)}{10^n Gamma(n+2)Gammaleft(frac{4}{3}right)}=frac{3sqrt{3}}{2pi}sum_{ngeq 1}left(tfrac{3}{10}right)^n Bleft(tfrac{2}{3},n+tfrac{4}{3}right) $$
where
$$ sum_{ngeq 1}left(tfrac{3}{10}right)^n Bleft(tfrac{2}{3},n+tfrac{4}{3}right) = int_{0}^{1}sum_{ngeq 1}left(tfrac{3}{10}right)^n(1-x)^{-1/3}x^{n+1/3},dx=int_{0}^{1}frac{3x^{4/3},dx}{(1-x)^{1/3}(10-3x)} $$
and the last integral can be computed in a explicit way with a bit of patience. The final outcome is
$$sum_{ngeq 1}frac{prod_{k=1}^{n}(3k+1)}{10^n(n+1)!}=color{red}{10sqrt[3]{frac{10}{7}}-11} $$
which can also be proved by invoking Lagrange's inversion theorem or the extended binomial theorem.

$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
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},
{Gammapars{4/3 + bracks{n - 1}}/Gammapars{4/3} over n!}
\[5mm] = &
{10 over 3},{pars{-2/3}! over Gammapars{4/3}}
sum_{n = 2}^{infty}pars{3 over 10}^{n},
{pars{n - 2/3}! over n!pars{-2/3}!}
\[5mm] = &
{10 over 3},{Gammapars{1/3} over pars{1/3}Gammapars{1/3}}
sum_{n = 2}^{infty}pars{3 over 10}^{n},{n - 2/3 choose n}
\[5mm] = &
10sum_{n = 2}^{infty}pars{3 over 10}^{n}
bracks{{-1/3 choose n}pars{-1}^{n}}
\[5mm] = &
10bracks{%
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] = &
10braces{bracks{1 + pars{-,{3 over 10}}}^{-1/3} - 1 - {1 over 10}}
=
bbx{10pars{10 over 7}^{1/3} - 11}
\[5mm] approx & 0.2625
end{align}