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: Note that $$\sum _{ n=m }^{ \infty  } (1/2)^{ n }=\frac { 1 }{ { 2 }^{ m } } +\frac { 1 }{ { 2 }^{ m+1 } } +...=\frac { 1 }{ { 2 }^{ m } } \left( 1+\frac { 1 }{ 2 } +... \right) =\frac { 1 }{ { 2 }^{ m } } \frac { 1 }{ 1-\frac { 1 }{ 2 }  } =\frac { 1 }{ { 2 }^{ m-1 } } $$
A: 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.
A: It may be worthwhile to think of the geometric series formula without the indices:  If $|r|<1$, then $$a+ar+ar^2+\cdots=\dfrac{a}{1-r}=\frac{\text{first term}}{1-\text{ratio}}.$$  
You can recognize a geometric series whenever you multiply by the same number to go from one term to the next, and that is your "$r$" (the ratio of successive terms).  Regardless of how you are indexing, the "$a$" is just the first term.  In your case, $a=\dfrac{1}{2^m}$.
