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


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 $z=1/8$.
