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
$$S=8\,\Biggl(1-\frac12-f(1)\Biggr)=4\,.$$
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}[1]{\mathrm{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{{4 \over 10} + {4 \cdot 7 \over 10 \cdot 20} +
{4 \cdot 7 \cdot 10 \over 10 \cdot 20 \cdot 30} + \cdots =
\sum_{n = 1}^{\infty}{\prod_{k = 1}^{n}\pars{3k + 1} \over
\prod_{k = 1}^{n}10k}:\ {\LARGE ?}}$.
\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
Good so far, to finish the proof notice that,
$$\frac{1}{1-z} = \sum_{i=0}^{\infty} z^i $$
Differentiating,
$$\frac{1}{(1-z)^2} = \sum_{i=1}^{\infty} i z^{i-1}$$
Multiply by $z$ then differentiate again,
$$z\frac{d}{dz} \frac{z}{(1-z)^2} = \sum_{i=1}^\infty i^2 z^{i} $$
So we have that,
$$\frac{z(z+1)}{(1-z)^3} = \sum_{k=1}^\infty k^2 z^k $$
Put in $z = x^2$ to obtain,
$$\sum_{k=1}^\infty k^2x^{2k} = \frac{x^2(x^2+1)}{(1-x^2)^3}$$
The most brute force way to calculate the integral after that is to substitute in $x = \sin \theta$, expand everything and separate and calculate all the integrals separately.