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? 

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 $(1)$.
