Summation of Central Binomial Coefficients divided by even powers of $2$ Whilst working out this problem the following summation emerged:
$$\sum_{m=0}^n\frac 1{2^{2m}}\binom {2m}m$$
The is equivalent to 
$$\begin{align}
\sum_{m=0}^n \frac {(2m-1)!!}{2m!!}&=\frac 12+\frac {1\cdot3}{2\cdot 4}+\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}+\cdots +\frac{1\cdot 3\cdot 5\cdot \cdots \cdot(2n-1)}{2\cdot 4\cdot 6\cdot \cdots \cdot 2n}\\
&=\frac 12\left(1+\frac 34\left(1+\frac 56\left(1+\cdots \left(1+\frac {2n-1}{2n}\right)\right)\right)\right)
\end{align}$$
 and terms are the same as coefficients in the expansion of $(1-x)^{-1/2}$.
Once the solution 
$$ \frac {n+1}{2^{2n+1}}\binom {2n+2}{n+1}$$
is known, the telescoping sum can be easily derived, i.e. 
  $$\frac 1{2^{2m}}\binom {2m}m=\frac {m+1}{2^{2(m+1)-1}}\binom {2(m+1)}{m+1}-\frac m{2^{2m-1}}\binom {2m}m$$
However, without knowing this a priori, how would we have approached this problem? 

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 ($\int\frac{dx}{\sin^2 x}=-\cot x$).
