Does the series $ \sum_{i=1}^{\infty}\frac{2^{i-1}(i-1)!}{\prod_{j=1}^{i}(2j+1)}$ converge and if so, to what? I am trying to prove the following series converges and find its limit if indeed it does:
$$ \sum_{i=1}^{\infty}\frac{2^{i-1}(i-1)!}{\prod_{j=1}^{i}(2j+1)}$$
Intuitively, I can see this is converging since the $i$-term product on the denominator, when expanded, should yield something at least as great as $2^i$ as a result of the $2$ coefficient, as well as something at least as great as the $(i-1)!$ due to $j$ itself.  However, I am not sure how to formally prove this.  Some preliminary calculations show this may converge to something close to but less than $1$ (maybe $1$ itself?) but how can I obtain the exact limit?
 A: Simplifying the Sum
$$
\begin{align}
\sum_{n=1}^\infty\frac{(n-1)!\,2^{n-1}}{\prod_{k=1}^n(2k+1)}
&=\sum_{n=1}^\infty\frac{(n-1)!\,2^{n-1}}{\frac{(2n+1)!}{n!\,2^n}}\\
&=\sum_{n=1}^\infty\frac1{2n(2n+1)}\frac{n!\,2^n}{\frac{(2n)!}{n!\,2^n}}\\
&=\sum_{n=1}^\infty\frac1{2n(2n+1)}\frac{4^n}{\binom{2n}{n}}\tag{1}
\end{align}
$$
Since $\binom{2n}{n}\sim\frac{4^n}{\sqrt{\pi n}}$, we have $\frac1{2n(2n+1)}\frac{4^n}{\binom{2n}{n}}\sim\frac{\sqrt\pi}{4n^{3/2}}$ and the sum converges since $\frac32\gt1$.

Evaluation $\boldsymbol{1}$
Equation $(2)$ from this answer says
$$
\sum_{n=1}^\infty\frac{4^nx^{2n}}{\binom{2n}{n}}
=\frac1{1-x^2}\left[x^2+\frac{x}{\sqrt{1-x^2}}\sin^{-1}(x)\right]\tag{2}
$$
Dividing $(2)$ by $x$ and integrating twice gives
$$
\sum_{n=1}^\infty\frac1{2n(2n+1)}\frac{4^nx^{2n+1}}{\binom{2n}{n}}
=x-\sqrt{1-x^2}\sin^{-1}(x)\tag{3}
$$
Evaluating $(3)$ at $x=1$ gives
$$
\sum_{n=1}^\infty\frac{(n-1)!\,2^{n-1}}{\prod_{k=1}^n(2k+1)}=1\tag{4}
$$

Evaluation $\boldsymbol{2}$
Using the Beta Function integral, we get
$$
\frac1{\binom{2n}{n}}=(2n+1)\int_0^1x^n(1-x)^n\,\mathrm{d}x\tag{5}
$$
Therefore,
$$
\begin{align}
\sum_{n=1}^\infty\frac1{2n(2n+1)}\frac{4^n}{\binom{2n}{n}}
&=\int_0^1\sum_{n=1}^\infty\frac{4^n}{2n}x^n(1-x)^n\,\mathrm{d}x\tag{6a}\\
&=-\frac12\int_0^1\log(1-4x(1-x))\,\mathrm{d}x\tag{6b}\\
&=-\int_0^1\log|1-2x|\,\mathrm{d}x\tag{6c}\\
&=-\frac12\int_{-1}^1\log|x|\,\mathrm{d}x\tag{6d}\\
&=-\int_0^1\log(x)\,\mathrm{d}x\tag{6e}\\[6pt]
&=1\tag{6f}
\end{align}
$$
Explanation:
$\text{(6a)}$: apply $(5)$
$\text{(6b)}$: use the Taylor series for $\log(1-x)$
$\text{(6c)}$: $\sqrt{1-4x+4x^2}=|1-2x|$
$\text{(6d)}$: substitute $x\mapsto\frac{1-x}2$
$\text{(6e)}$: symmetry
$\text{(6f)}$: $\int\log(x)\,\mathrm{d}x=x\log(x)-x$
A: $\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

\begin{equation}
\sum_{i = 1}^{\infty}{2^{i - 1}\pars{i - 1}! \over
\prod_{j = 1}^{i}\pars{2j + 1}}:\ ?\label{1}\tag{1}
\end{equation}

\begin{align}
&\bbox[8px,border:0.1em groove navy]{{2^{i - 1}\pars{i - 1}! \over \prod_{j = 1}^{i}\pars{2j + 1}}} =
{2^{i - 1}\pars{i - 1}! \over 2^{i}\prod_{j = 1}^{i}\pars{j + 1/2}}
\\[5mm] = &\
{1 \over 2}\,{\pars{i - 1}! \over \pars{3/2}_{i}}
\qquad\pars{~\vphantom{\large A}\pars{a}_{n}:\ Pochhammer\ Symbol~}
\\[5mm] = &\
{1 \over 2}\,{\pars{i - 1}! \over \Gamma\pars{3/2 + i}/\Gamma\pars{3/2}} =
{1 \over 2}\,\
\underbrace{{\Gamma\pars{i}\Gamma\pars{3/2} \over \Gamma\pars{i + 3/2}}}
_{\ds{\mrm{B}\pars{i,3/2}}}\qquad
\pars{~\mrm{B}:\ Beta\ Function~}
\\ = &\,\,
\bbox[8px,border:0.1em groove navy]{{1 \over 2}
\int_{0}^{1}x^{i - 1}\pars{1 - x}^{1/2}\,\dd x}\label{2}\tag{2}
\end{align}

Note that
  $\ds{\pars{a}_{n} = {\Gamma\pars{a + n} \over \Gamma\pars{a}}}$ where
  $\ds{\Gamma}$ is the Gamma Function. By inserting the result \eqref{2} into the expression \eqref{1}:


\begin{align}
\color{#f00}{\sum_{i = 1}^{\infty}{2^{i - 1}\pars{i - 1}! \over
\prod_{j = 1}^{i}\pars{2j + 1}}} & =
\sum_{i = 1}^{\infty}{1 \over 2}
\int_{0}^{1}x^{i - 1}\pars{1 - x}^{1/2}\,\dd x =
{1 \over 2}\int_{0}^{1}{\dd x \over \root{1 - x}} = \color{#f00}{1}
\end{align}
A: $$2^{i-1}(i-1)! = \prod_{j=1}^{i-1} (2j)$$
so that
$$
\frac{2^{i-1}(i-1)!}{\prod_{j=1}^{i} (2j+1)}
= \frac{1}{2i+1}\prod_{j=1}^{i-1} \frac{2j}{2j+1}
= \frac{1}{2i+1}\prod_{j=1}^{i-1} \left(1-\frac{1}{2j+1}\right)
$$
Now, how does the general term
$$
a_i \stackrel{\rm def}{=} \frac{1}{2i+1}\prod_{j=1}^{i-1} \left(1-\frac{1}{2j+1}\right)
$$
of your series behave when $i\to\infty$?

Hint:
$$
\ln \prod_{j=1}^{i-1} \left(1-\frac{1}{2j+1}\right)
=  \sum_{j=1}^{i-1} \ln \left(1-\frac{1}{2j+1}\right) 
$$
and $\ln \left(1-\frac{1}{2j+1}\right) =_{j\to\infty} -\frac{1}{2j+1} + O\left(\frac{1}{j^2}\right)$ so that, by theorems of comparison (for series with terms of constant sign)
$$
\ln \prod_{j=1}^{i-1} \left(1-\frac{1}{2j+1}\right)
=  -\sum_{j=1}^{i-1} \frac{1}{2j+1} + O(1)
= \frac{1}{2}\ln i + o(\ln i)
$$
and then $a_i = \frac{1}{2i+1}\cdot \frac{1}{\sqrt{i}}\cdot f(i)$ for $f(i) = e^{o(\ln i)}$. This suffices to prove convergence, by comparison to a $p$-series (for say, $1<p<3/2$).
