Series sum with factorial notation The sum of series 
$\displaystyle 1+\frac{1}{1!}\cdot \frac{1}{4}+\frac{1\cdot 3}{2!}\cdot \frac{1}{4^2}+\frac{1\cdot 3 \cdot 5}{3!}\cdot \frac{1}{4^3}+\cdots $
what i try:
$$1+\lim_{n\rightarrow \infty}\sum^{n}_{r=1}\frac{1\cdot 3\cdot \cdots (2r-1)}{r!\cdot 4^r}$$
$$1+\lim_{n\rightarrow \infty}\sum^{n}_{r=1}\frac{(2r)!}{r!\prod^{n}_{r=1}(2r)!}$$
How do i solve it Help me please
 A: 
The  series  under  consideration   is
\begin{align*}
\color{blue}{1+\sum_{r=1}^\infty\frac{(2r-1)!!}{r!4^r}}&=1+\sum_{r=1}^\infty\frac{(2r)!}{r!4^r(2r)!!}\\
&=1+\sum_{r=1}^\infty\frac{(2r)!}{r!r!}\left(\frac{1}{8}\right)^r\\
&=\sum_{r=0}^\infty\binom{2r}{r}\left(\frac{1}{8}\right)^r\\
&=\left.\frac{1}{\sqrt{1-4z}}\right|_{z=1/8}\tag{1}\\
&=\frac{1}{\sqrt{1-\frac{1}{2}}}\\
&\,\,\color{blue}{=\sqrt{2}}
\end{align*}

In (1) we use the ordinary generating function of the central binomial coefficients evaluated at $z=1/8$.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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\begin{align}
&\bbox[10px,#ffd]{1 + {1 \over 1!}\,{1 \over 4} +
{1\cdot 3 \over 2!}\, {1 \over 4^{2}} +
{1\cdot 3 \cdot 5 \over 3!}\,{1 \over 4^{3}} + \cdots} =
1 + \sum_{n = 1}^{\infty}{\prod_{k = 0}^{n - 1}\pars{2k + 1} \over
n!\, 4^{n}}
\\[5mm] = &\
1 + \sum_{n = 1}^{\infty}{2^{n}\prod_{k = 0}^{n - 1}\pars{k + 1/2} \over
n!\, 4^{n}} =
1 + \sum_{n = 1}^{\infty}{\pars{1/2}^{\overline{n}} \over n!}
\,\pars{1 \over 2}^{n}
\\[5mm] = &\
1 + \sum_{n = 1}^{\infty}{\Gamma\pars{1/2 + n}/\Gamma\pars{1/2} \over n!}
\,\pars{1 \over 2}^{n} =
\sum_{n = 0}^{\infty}{\pars{n - 1/2}! \over n!\pars{-1/2}!}
\,\pars{1 \over 2}^{n}
\\[5mm] = &\
\sum_{n = 0}^{\infty}{n - 1/2 \choose n}\,\pars{1 \over 2}^{n} =
\sum_{n = 0}^{\infty}{-1/2 \choose n}\,\pars{-\,{1 \over 2}}^{n} =
\bracks{1 + \pars{-1 \over 2}}^{-1/2}
\\[5mm] = &\ \bbx{\root{2}} \approx 1.4142
\end{align}
