How do I find the partial sum of this So I have a sum defined below:

$$
\sum_{m=1}^n 2^{-m}
$$

I know the partial sum equals 

$$
\frac{1}{2^n}(2^n - 1)\ 
$$

But how do you go from one to the other?
Let 
$$S=\sum_{k=1}^n 2^{-k}=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\cdots +\frac{1}{2^n}.\tag{1}$$
Then 
$$2S=1+\frac{1}{2}+\frac{1}{2^2}+\cdots+\frac{1}{2^{n-1}}.\tag{2}$$
From (1) and (2), by subtraction we obtain
$$S=1-\frac{1}{2^n}$$
(note the massive amount of cancellation). Now we are almost at the desired expression.
By Taylor series you can find $\forall x : |x|&lt;1$

$$\frac{1}{1-x}=\sum_{k=0}^{\infty} x^k$$

And so

$$\sum_{k=0}^{m} x^k=\sum_{k=0}^{\infty} x^k - \sum_{k=m+1}^{\infty} x^k=(1-x^{m+1})\sum_{k=0}^{\infty} x^k $$

$$=\frac{1-x^{m+1}}{1-x}$$