You almost got everything right, and the only problem you have is a minor error when you define $b_n$. You should have
$$b_n=2^n\,(n+1)!=2^n\,\Gamma(n+2)\text{ for }n=1,2,3,\ldots\,.$$
Thus, for each $n=1,2,3,\ldots$,
$$\begin{align}\frac{a_n}{b_n}&=\frac{\Gamma\left(n+\frac12\right)}{\Gamma\left(\frac12\right)\,\Gamma(n+2)}=\frac{2}{\pi}\,\left(\frac{\Gamma\left(n+\frac12\right)\,\Gamma\left(\frac32\right)}{\Gamma(n+2)}\right)\\&=\frac{2}{\pi}\,\text{B}\left(n+\frac12,\frac32\right)\,,\end{align}$$
where $\Gamma$ and $\text{B}$ are the usual gamma and beta functions, respectively. Hence,
$$\frac{a_n}{b_n}=\frac{2}{\pi}\,\int_0^1\,x^{n-\frac12}\,(1-x)^{\frac12}\,\text{d}x\,,$$
so
$$\begin{align}\sum_{n=1}^\infty\,\frac{a_n}{b_n}&=\frac{2}{\pi}\,\int_0^1\,\frac{x^{\frac12}}{1-x}\,(1-x)^{\frac12}\,\text{d}x
\\&=\frac{2}{\pi}\,\int_0^1\,x^{\frac12}\,(1-x)^{-\frac12}\,\text{d}x\,.\end{align}$$
That is, with $u:=x^{\frac12}$, we obtain
$$\begin{align}\sum_{n=1}^\infty\,\frac{a_n}{b_n}&=\frac{4}{\pi}\,\int_0^1\,\frac{u^2}{\sqrt{1-u^2}}\,\text{d}u\\&=\frac{2}{\pi}\,\left(\text{arcsin}(u)-u\,\sqrt{1-u^2}\right)\Big|_{u=0}^{u=1}\,.\end{align}$$
Ergo,
$$\sum_{n=1}^\infty\,\frac{a_n}{b_n}=1\,,$$
whence
$$1+\frac{1\cdot 3}{6}+\frac{1\cdot 3\cdot 5}{6\cdot 8}+\frac{1\cdot3\cdot 5\cdot 7}{6\cdot 8\cdot 10}+\ldots=4\,\sum_{n=1}^\infty\,\frac{a_n}{b_n}=4\,.$$
In fact, one can show that
$$f(z):=\sum_{n=0}^\infty\,\prod_{k=1}^n\,\left(\frac{k-\frac32}{k}\right)\,z^n=(1-z)^{\frac12}$$
for all $z\in\mathbb{C}$ with $|z|\leq 1$. The requested sum satisfies
$$S=8\,\Biggl(1-\frac12-f(1)\Biggr)=4\,.$$
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}[1]{\mathrm{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
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$\ds{{4 \over 10} + {4 \cdot 7 \over 10 \cdot 20} +
{4 \cdot 7 \cdot 10 \over 10 \cdot 20 \cdot 30} + \cdots =
\sum_{n = 1}^{\infty}{\prod_{k = 1}^{n}\pars{3k + 1} \over
\prod_{k = 1}^{n}10k}:\ {\LARGE ?}}$.
\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}
Best Answer
Such integrals can be expressed by the digamma function. One of the formulae for the digamma function is
$$ \psi(z+1) = -\gamma + \int_0^1 \frac{1-t^z}{1-t}dt $$
You have then $$ \int_0^1 \frac{1-x^{3n}}{1-x^3} dx = \int_0^1 \frac{1-t^{n}}{1-t} \frac13t^{-\frac23}dt = \int_0^1 \frac13 \Big(\frac{1-t^{n-\frac23}}{1-t}-\frac{1-t^{-\frac23}}{1-t}\Big)dt = \frac13\big(\psi(n+\frac13)-\psi(\frac13)\big)$$ $$ \int_0^1 \frac{x^2(1-x^{2n)}}{1-x^2} dx = \int_0^1 \frac{t(1-t^{n})}{1-t} \frac12t^{-\frac12}dt = \int_0^1 \frac12 \Big(2\frac{1-t^{n+\frac12}}{1-t}-2\frac{1-t^{\frac12}}{1-t}\Big)dt = \frac12\big(\psi(n+\frac32)-\psi(\frac32)\big)$$ so $$ \sum_{k=1}^n \frac{1}{(3k-2)(2k+1)} = \frac17\big(\psi(n+\frac13)-\psi(n+\frac32)-\psi(\frac13)+\psi(\frac32)\big)$$ The same formula can be derived faster, using the fact that $$ \psi(n+z)-\psi(z) = \sum_{k=0}^{n-1}\frac{1}{z+k}$$ so $$ \sum_{k=1}^{n}\frac{1}{k-\frac23}=\sum_{k=0}^{n-1}\frac{1}{k+\frac13} = \psi(n+\frac13)-\psi(\frac13) $$ $$ \sum_{k=1}^{n}\frac{1}{k+\frac12}=\sum_{k=0}^{n-1}\frac{1}{k+\frac32} = \psi(n+\frac32)-\psi(\frac32) $$
If you want to calculate the infinite sum, you have \begin{align} \sum_{k=1}^\infty \frac{1}{(3k-2)(2k+1)} &= \frac17 \int_0^1 \Big(\frac{3}{1-x^3} - \frac{2x^2}{1-x^2}\Big) dx =\\ &= \frac17\int_0^1 \Big(\big(\frac{1}{1-x}+\frac{x+2}{x^2+x+1}\big) - \big(-2 + \frac{1}{1-x}+\frac{1}{x+1}\big)\Big) dx =\\ &= \frac17 \int_0^1 \Big(\frac{x+2}{x^2+x+1} +2 - \frac{1}{x+1}\Big) dx \end{align} and the last integral can be computed with standard methods for the integrals of rational functions:
\begin{align} \int_0^1 \frac{x+2}{x^2+x+1} dx &= \int_0^1 \frac{(x+\frac12)+\frac32}{(x+\frac12)^2+\frac34} dx =^{t=\frac{2}{\sqrt{3}}(x+\frac12)} \\ &= \int_{1/\sqrt{3}}^{\sqrt{3}} \frac{t+\sqrt{3}}{t^2+1} dt = \\ &= \Big(\frac{1}{2}\log(t^2+1) + \sqrt{3}\arctan t\Big)\Big|_{t=1/\sqrt{3}}^{t=\sqrt{3}} = \\ &= \frac{1}{2}\log 3 + \frac{\pi\sqrt{3}}{6}\end{align} $$ \int_0^1\frac{1}{x+1} dx = \log(x+1)\big|_{x=0}^{x=1}= \log 2 $$ so $$ \sum_{k=1}^\infty \frac{1}{(3k-2)(2k+1)} =\frac17 (2+ \frac{\pi\sqrt{3}}{6}+\frac{1}{2}\log 3 -\log 2) $$