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 $S = P?$
 A: Note that setting $nx=\frac{3}{4}$ and $\frac{n(n-1)}{2!}=\frac{3\cdot 5}{4\cdot 8}$ in order to get values for $n$  and $x$ is just a clever approach, but not a proof that $P=S$.
To make it a proof we have to additionally verify that the general terms evaluated at $n=\frac{3}{2}$ and $x=-\frac{1}{2}$ are equal.

Let's write $S$ and $P$ with general term. We have
\begin{align*}
S&=1+\frac{3}{4}+\frac{3\cdot 5}{4\cdot 8}+\cdots+
\color{blue}{\frac{3\cdot 5\cdots (2k+1)}{4\cdot 8\cdots (4k)}}+\cdots\tag{1}\\
P&=(1+x)^n\\
&=1+nx+\frac{n(n-1)}{2!}x^2+\cdots+\color{blue}{\frac{n(n-1)\cdots(n-k+1)}{k!}x^k}+\cdots\tag{2}
\end{align*}
We simplify the general term of $S$ somewhat
\begin{align*}
\frac{3\cdot 5\cdots (2k+1)}{4\cdot 8\cdots (4k)}&=\frac{1}{4^k}\cdot\frac{3\cdot 5\cdots (2k+1)}{1\cdot 2\cdots k}\\
&\,\,\color{blue}{=\frac{1}{4^kk!}\prod_{j=1}^k(2j+1)}
\end{align*}
The general term of $P$ evaluated $n=\frac{3}{2}$ and $x=-\frac{1}{2}$:
\begin{align*}
&\left.\frac{n(n-1)\cdots(n-k+1)}{k!}x^k\right|_{n=-\frac{3}{2},x=-\frac{1}{2}}\\
&\qquad\qquad=\frac{1}{k!}\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)\cdots\left(-\frac{3}{2}-k+1\right)\left(-\frac{1}{2}\right)^k\\
&\qquad\qquad=\frac{1}{k!}\frac{(-1)^k}{2^k}\left(3\right)\left(5\right)\cdots\left(3+2k-2\right)\left(-\frac{1}{2}\right)^k\\
&\qquad\qquad\,\,\color{blue}{=\frac{1}{4^kk!}\prod_{j=1}^k(2j+1)}
\end{align*}
Since both terms are equal we can now conclude that $S=P$.

Note: In order to generalise this approach,  we could start to study and compare general terms accordingly.
[Add-on (2020-07-08)]: This add-on is based on OP who has successfully calculated the generalisation in comments to this answer.
The general ($k$-th) term of $S = 1 + \sum_{n=1}^\infty\prod_{j=1}^n\frac{aj+b}{dj}$ is (provided $a\ne 0$):
\begin{align*}
\prod_{j=1}^k\frac{aj+b}{dj}&=\frac{1}{d^kk!}\prod_{j=1}^k(aj+b)\\
&=\left(\frac{a}{d}\right)^k\frac{1}{k!}\prod_{j=1}^k\left(j+\frac{b}{a}\right)\tag{3}
\end{align*}
Since the general term of $P=(1+x)^n$ is
\begin{align*}
\frac{x^k}{k!}\prod_{j=1}^k(n-j+1)&=\frac{(-x)^k}{k!}\prod_{j=1}^{k}\left(j-\left(n+1\right)\right)\tag{4}
\end{align*}
we obtain by comparison of (3) with  (4):
\begin{align*}
x=-\frac{a}{d}\qquad\qquad n=-\left(1+\frac{b}{a}\right)
\end{align*}

We conclude, providing $|x|=\left|\frac{a}{d}\right|<1$ to assure convergence of the binomial series:
\begin{align*}
\color{blue}{S }= 1 + \sum_{n=1}^\infty\prod_{j=1}^n\frac{aj+b}{dj}\color{blue}{=\left(1-\frac{a}{d}\right)^{-\left(1+\frac{b}{a}\right)}}
\end{align*}

Hints:

*

*Alternatively to OPs generalisation we can also recall that $P=(1+x)^n$ is some kind of reference series used to derive $S$. We can turn the tables and play with different settings of $x$ and $n$ (respecting $|x|<1$) and check which different general terms and series $S$ we obtain this way.


*A technical aspect. We have to be aware of arithmetic precedence rules and write $\prod_{j=1}^k\color{blue}{(}aj+b\color{blue}{)}$ using brackets, since  we have
\begin{align*}
\prod_{j=1}^kaj+b&=\left(\prod_{j=1}^kaj\right)+b=a^kk!+b\\
\prod_{j=1}^kb+aj&=\left(\prod_{j=1}^kb\right)+aj=b^k+aj\\
\prod_{j=1}^k\left(aj+b\right)&=(a+b)(2a+b)\cdots(ka+b)
\end{align*}
You might want to see this answer for more information regarding precedence rules.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \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}\,}\,}
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\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}
