Summation of Central Binomial Coefficients divided by even powers of $2$
You may do the following to get rid of the powers of two and transform it into a standard sum: \begin{equation} \binom{2m}{m}=\frac{(2m)!}{m!m!}=\frac{2^m m! 1\cdot3\cdot5\cdot \cdots (2m-1)}{m!m!}=2^{2m} \frac{(1/2)(3/2)\cdots (m-1/2)}{m!}=2^{2m}\binom{m-1/2}{m}~~~~~~ (1) \end{equation} Then your sum becomes $$\sum_{m=0}^{n} \binom{m-1/2}{m}=\binom{n-1/2 +1}{n} =\binom{n+1-1/2}{n}=\frac{n+1}{n+1-1/2-n}\frac{n+1-1/2-n}{n+1}\binom{n+1-1/2}{n}=\frac{n+1}{1/2}\binom{n+1-1/2}{n+1}=2(n+1)\frac{1}{2^{2(n+1)}}\binom{2(n+1)}{n+1}=\frac{n+1}{2^{2{n+1}}}\binom{2n+2}{n+1},$$ where the first equality is just the usual identity $\sum_{j=0}^n \binom{r+j}{j} =\binom{r+n+1}{n}$ valid for any real $r$ and any non-negative integer $n$, and the last few steps were just trying to slightly transform the coefficient in a way that allowed application of $(1)$.
Another approach. By Euler/De Moivre's formula we have $$A_n=\frac{1}{4^n}\binom{2n}{n} = \frac{1}{2\pi}\int_{-\pi}^{\pi}\cos^{2n}(x)\,dx \tag{1}$$
since $\cos(x)^{2n} = \frac{1}{4^n}\sum_{j=0}^{2n}\binom{2n}{j}e^{(2n-j)ix}e^{-jix}$ and $\int_{-\pi}^{\pi}e^{kix}\,dx =2\pi \delta(k)$, hence only the contribute given by $j=n$ survives. It follows that $$ \color{red}{\sum_{n=0}^{N}\frac{1}{4^n}\binom{2n}{n}}=\frac{1}{2\pi}\int_{-\pi}^{\pi}\frac{1-\cos^{2N+2}(x)}{\sin^2(x)}\,dx =(2N+2)\,A_{N+1}=\color{red}{\frac{N+1}{2^{2N+1}}\binom{2N+2}{N+1}}\tag{2}$$ by integration by parts ($\int\frac{dx}{\sin^2 x}=-\cot x$).
I've already accepted Sebastian Munoz's very good solution above.
This is a paraphrase of his solution for my own reference.
$$\begin{align} \sum_{m=0}^n\frac 1{2^{2m}}\binom {2m}m &\color{lightgrey}{=\sum_{m=0}^n\frac 1{2^{2m}}\frac {(2m)!}{m!m!}}\\ &\color{lightgrey}{=\sum_{m=0}^n\frac 1{2^{2m}}\frac{(2m)(2m-1)(2m-2)\cdots 3\cdot 2\cdot1}{(1\cdot 2\cdot3\cdot\cdots \cdot m)(1\cdot 2\cdot3\cdot\cdots \cdot m)}}\\ &\color{lightgrey}{=\sum_{m=0}^n \underbrace{\frac {(2m)(2m-2)(2m-4)\cdots 2}{2^m(1\cdot 2\cdot3\cdot\cdots \cdot m)}}_{=1}\cdot \frac{(2m-1)(2m-3)(2m-5)\cdots3\cdot 1}{2^{m}\cdot (1\cdot 2\cdot3\cdot\cdots \cdot m)}}\\ &\color{lightgrey}{=\sum_{m=0}^n\frac{\frac{2m-1}2\cdot \frac {2m-3}2\cdot \frac {2m-5}2\cdots \frac 32\cdot \frac 12}{1\cdot 2\cdot3\cdot\cdots \cdot m}}\\ &\color{lightgrey}{=\sum_{m=0}^n\frac{\left(m-\frac 12\right)\left(m-\frac 32\right)\left(m-\frac 52\right)\cdots \frac 32\cdot \frac 12}{1\cdot 2\cdot3\cdot\cdots \cdot m}}\\ &=\sum_{m=0}^n \binom {m-\frac 12}m\\ &=\binom {n+\frac 12}n\\ &\color{lightgrey}{=\frac {(n+\frac12)(n-\frac 12)(n-\frac 32)\cdots \frac 32}{1\cdot2 \cdot3 \cdot\quad\cdots \cdot n\qquad}}\\ &\color{lightgrey}{=\frac 1{2^n}\cdot \frac {(2n+1)(2n-1)(2n-3)\cdots 3}{1\cdot 2\cdot 3\cdot\cdots\cdot n}}\color{lightblue}{\cdot \frac{(2n+2)(2n)(2n-2)\cdots 2}{2^{n+1}(n+1)(n)(n-1)\cdots 1}\cdot \frac{n+1}{n+1}}\\ &=\frac {n+1}{2^{2n+1}}\cdot \frac{(2n+2)!}{(n+1)!(n+1)!}\\ &=\frac {n+1}{2^{2n+1}}\binom {2n+2}{n+1}\quad\blacksquare\end{align}$$
Additional note: (added Oct 2018)
Note that $$\begin{align} \frac 1{\sqrt{1-4x}} &=(1-4x)^{-\frac 12}\\ &=\sum_{r=0}^\infty \binom {-\frac 12}r (-4x)^r = \sum_{r=0}^\infty \binom {-\frac 12}r (-1)^4 r^4x^r\tag{*} \\\ &=\sum_{r=0}^\infty \frac {\left(-\dfrac 12\right)_r}{r!} (-4x)^r\\ &=\sum_{r=0}^\infty \frac {(-1)^r(2r-1)!!}{(2^r)}\cdot \frac 1{r!}\cdot (-1)^r4^rx^r\\ &=\sum_{r=0}^\infty \frac {(2r-1)!!2^r}{r!}x^r\cdot \color{lightgrey}{\frac {r!}{r!}}\\ &=\sum_{r=0}^\infty \frac {(2r-1)!!(2r)!!}{r!r!}x^r\\ &=\sum_{r=0}^\infty\frac {(2r)!}{r!r!} x^r\\ &=\sum_{r=0}^\infty\binom {2r}r x^r\tag{**} \end{align}$$
From (*) and (**),
$$\begin{align} (-1)^r 4^r\binom {-\frac 12}r&=\qquad\;\binom {2r}r\\ \binom {-\frac 12}r&=\frac {(-1)^r}{4^r}\binom {2r}r\\ (-1)^r\binom {r-\frac 12}r&=\frac {(-1)^r}{4^r}\binom {2r}r\\ \binom {r-\frac 12}r&=\frac 1{\;2^{2r}}\;\binom {2r}r\end{align}$$