An alternative way to find the sum of this series?

Question

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

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?

Answer 1

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.

Answer 2

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