Summation Algebra: $\sum_{n=2}^\infty \frac{1}{2^{n-1}}$ What rule do we use to get the numerical value of $$\sum_{n=2}^\infty \frac{1}{2^{n-1}}$$ 
My notes give the answer straight away, without explaining what the steps are.
// Thanks everyone! 

$$\sum_{n=2}^\infty \frac{1}{2^{n-1}}=\frac{1}{2}\sum_{n=2}^\infty \frac{1}{2^{n-2}}=\frac{1}{2}\left(1+\frac{1}{2}+\frac{1}{4}+...\right)=\frac{1}{2}\frac{1}{1-\frac{1}{2}}=1$$ 
