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 $1/(1-0.5)=2$.

Note that $$\sum _{ n=m }^{ \infty  } (1/2)^{ n }=\frac { 1 }{ { 2 }^{ m } } +\frac { 1 }{ { 2 }^{ m+1 } } +...=\frac { 1 }{ { 2 }^{ m } } \left( 1+\frac { 1 }{ 2 } +... \right) =\frac { 1 }{ { 2 }^{ m } } \frac { 1 }{ 1-\frac { 1 }{ 2 }  } =\frac { 1 }{ { 2 }^{ m-1 } } $$
