What is the sum of a geometric series indexed from $m>0$? I know how to calculate the sum of the geometric series given the base is $|x|<1.$
But how would I calculate something like $$\sum_{n=m}^{\infty}(1/2)^n$$
If it was from zero to infinity, the answer would be $2$ because $1/(1-0.5)=2$.

A slightly different approach is to split up your sum into known sums.  First notice the sum you're interested in is:
$$ \sum_{n=m}^\infty \frac{1}{2^n} = \sum_{n=0}^\infty \frac{1}{2^n}-\sum_{n=0}^m \frac{1}{2^n}$$  The first sum is just a geometric series, the other is just a finite geometric series which is also well know see here:
http://mathworld.wolfram.com/GeometricSeries.html
So we have $$\sum_{n=m}^\infty \frac{1}{2^n} = 2-\frac{1-(\frac{1}{2})^m}{1-\frac{1}{2}}=2^{1-m}$$  which is precisely what @Battani got as well.
