Sum of the inverse of a geometric series?

Question

I'm trying to solve for this summation:

$$\sum_{j=0}^{i} {\left(\frac 1 2\right)^j}$$

This looks a lot like a geometric series, but it appears to be inverted. Upon plugging the sum into Wolfram Alpha, I find the answer to be

$2-2^{-i}$

but I don't understand how it gets there. Am I able to consider this a geometric series at all? It almost seems closer to the harmonic series.

Answer 1

It is a geometric series:

$$\sum_{j=0}^{i} \frac{1}{2^j}=\sum_{j=0}^{i}\left( \frac{1}{2}\right)^j=\frac{1-\left(\frac12\right)^{i+1}}{1-\frac12}=2\left(1-\left(\frac12\right)^{i+1}\right)=2-\left(\frac12\right)^{i}=2-2^{-i}$$

Answer 2

As $\frac1{2^j}=\left(\frac12\right)^j$, this is also a geometric sum with the common ratio of $r=\frac12$. So you apply the formula for geometric sums $$\sum_{j=0}^nr^j=\frac{1-r^{n+1}}{1-r}$$to obtain the answer you have written.