What is the relation of the Infinite power series with these fraction series? Take this infinite series:
$$S = 1 + \sum_{n=1}^\infty\prod_{i=1}^n\frac{2i+1}{4i} = 1 + \frac{3}{4} + \frac{3\times5}{4\times8} + \frac{3\times5\times7}{4\times8\times12} + ....$$
We want to find the sum of this series. I didn't know how to solve this. But when I went to look at the solution, they compared this series with the infinite power series
$$P = (1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + ......$$
for some real $n$ and $x$. Equating the corresponding terms ($nx = \frac{3}{4}$ and $\frac{n(n-1)}{2}x^2 = \frac{3\times5}{4\times8}$) they found $n=-\frac{3}{2}$ and $x=-\frac{1}{2}$. So that the sum is simply $2^\frac{3}{2}$. And when I checked the infinite power series $P$ plugging these values of $x$ and $n$, it really turn out to be the series $S$. Now, I don't understand why this comparison works.
Let's generalise this thing. Say, $S$ is given by
$$S = 1 + \sum_{n=1}^\infty\prod_{i=1}^n\frac{ai+b}{di}$$
for some positive integers $a, b$ and $d$ with $b < a$, and it is guaranteed (given) that the series converges. Let $P$ be the same as above. Now, can anyone say if $S$, as defined just now, can always be compared with the series $P$, that is, are there always some real $n$ and $x$ such that $S = P?$

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
S & \equiv 1 + \sum_{n = 1}^{\infty}
\prod_{i = 1}^{n}{2i + 1 \over 4i} =
1 + \sum_{n = 1}^{\infty}{2^{n} \over 4^{n}}
{\prod_{i = 1}^{n}\pars{i + 1/2} \over n!} =
1 + \sum_{n = 1}^{\infty}{1 \over 2^{n}\, n!}\,
\pars{3 \over 2}^{\overline{\large n}}
\\[5mm] & =
1 + \sum_{n = 1}^{\infty}{1 \over 2^{n}\, n!}\,
{\Gamma\pars{3/2 + n} \over\Gamma\pars{3/2}} =
1 + \sum_{n = 1}^{\infty}{1 \over 2^{n}}\,
{\pars{n + 1/2}! \over n!\pars{1/2}!}
\\[5mm] & =
1 + \sum_{n = 1}^{\infty}{1 \over 2^{n}}\,{n + 1/2 \choose n} =
1 + \sum_{n = 1}^{\infty}{1 \over 2^{n}}
\bracks{{-3/2 \choose n}\pars{-1}^{n}}
\\[5mm] & =
1 + \sum_{n = 1}^{\infty}
{-3/2 \choose n}\pars{-\,{1 \over 2}}^{n} =
1 + \braces{\bracks{1 + \pars{-\,{1 \over 2}}}^{-3/2} - 1}
\\[5mm] & =
\bbox[15px,#ffd,border:1px solid navy]{\large 2\root{2}}\ \approx\ 2.8284
\end{align}
