Sum of series $1+\frac{1}{4}+\frac{1\times3}{4\times8}+\frac{1\times3\times5}{4\times8 \times12}+\cdots$ 
Sum of series :
  $$1+\frac{1}{4}+\frac{1\times3}{4\times8}+\frac{1\times3\times5}{4\times8 \times12}+\cdots$$


$n$-th term of series is 
$$a_{n} =\frac{1\times3 \times5\times\cdots \times(2n-1)}{4\times8\times12\times \cdots  \times4n} = \prod^{n}_{k=1}{2k-1\over4k}$$
I can not go further,
 A: Note, the general term $a_n, n\geq 1$ is
\begin{align*}
a_n=\frac{1\cdot 3\cdot 5\cdots (2n-1)}{4\cdot8\cdot 12\cdots (4n)}
=\frac{1\cdot 3\cdot 5\cdots (2n-1)}{(1\cdot 2\cdot 3\cdots n)\cdot 4^n}
=\frac{(2n-1)!!}{n!}\cdot \frac{1}{4^n}\\
\end{align*}
with $$(2n-1)!!=(2n-1)\cdot(2n-3)\cdots 5\cdot 3\cdot 1$$ double factorials.

We obtain
  \begin{align*}
1+\sum_{n=1}^\infty\frac{(2n-1)!!}{n!}\cdot\frac{1}{4^n}
&=\sum_{n=0}^\infty\frac{(2n)!}{n!(2n)!!}\cdot\frac{1}{4^n}\tag{1}\\
&=\sum_{n=0}^\infty\frac{(2n)!}{n!n!2^n}\cdot\frac{1}{4^n}\tag{2}\\
&=\sum_{n=0}^\infty\binom{2n}{n}\frac{1}{8^n}\tag{3}\\
&=\left.\frac{1}{\sqrt{1-4z}}\right|_{z=\frac{1}{8}}\tag{4}\\
&=\frac{1}{\sqrt{1-\frac{1}{2}}}\\
&=\sqrt{2}
\end{align*}

Comment:


*

*In (1) we apply $(2n)!=(2n)!!\cdot (2n-1)!!$.

*In (2) we use $(2n)!!=n!2^n$.

*In (3) we use $\binom{2n}{n}=\frac{(2n)!}{n!n!}$.

*In (4) we use the generating function of the central binomial coefficient.

Add-on [2017-03-12] according to a comment from @navinstudent.
Using the binomial series expansion we obtain
  \begin{align*}
\frac{1}{\sqrt{1-4z}}&=(1-4z)^{-\frac{1}{2}}=\sum_{n=0}^\infty \binom{-\frac{1}{2}}{n}(-4z)^n\qquad\qquad |z|<\frac{1}{4}
\end{align*}
  Since 
  \begin{align*}
\binom{-\frac{1}{2}}{n}&=\frac{\left(-\frac{1}{2}\right)\left(-\frac{1}{2}-1\right)\cdots\left(-\frac{1}{2}-n+1\right)}{n!}\\
&=\frac{1}{n!}\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\cdots\left(-\frac{2n-1}{2}\right)\\
&=\frac{(-1)^n}{2^nn!}\cdot(2n-1)!!
=\frac{(-1)^n}{2^nn!}\cdot\frac{(2n)!}{(2n)!!}
=\frac{(-1)^n}{2^nn!}\cdot\frac{(2n)!}{2^nn!}\\
&=\frac{(-1)^n}{4^n}\binom{2n}{n}
\end{align*}
  we get $$\sum_{n=0}^\infty \binom{-\frac{1}{2}}{n}(-4z)^n=\sum_{n=0}^\infty \binom{2n}{n}z^n$$ and step (4) follows.

