Not quite sure if I am summing up the geometric series correctly. I have the following series (and I think it is geometric).
$$\sum_{k=0}^{i-1} \frac{1}{2^k}$$
I am trying to find a closed form for it. Of course, assuming it is geometric? However the ratio changes each time, as I am adding a factor of $\frac{1}{2^k}$ which is not constant (depends on the k)
 A: You're confusing arithmetic and geometric series. In an arithmetic series, the difference between the successive terms is constant; in a geometric series, it's the ratio that stays the same. Your sum is a geometric series because the ratio between the $n$th and $(n+1)$th terms is $${{1\over 2^n}\over{1\over 2^{n+1}}}=2,$$ which does not depend on $n$.
There is of course a formula for summing geometric series' which you can just plug into. However, it's good to think about what's going on here: letting $I=\sum_{k=1}^{i-1}{1\over 2^k}$, we have $$I-{1\over 2}I={1\over 2}-{1\over 2^i}={2^{i-1}-1\over 2^i},\quad\mbox{so}\quad I={2^{i-1}-1\over 2^{i-1}}=1-{1\over 2^{i-1}}.$$ This is elementary, except for the first "$=$" which is a bit mysterious. Where it comes from, and the reason we look at "$I-{1\over 2}I$", is because ${1\over 2}$ is the ratio between the terms so a bunch of terms cancel out in $I-{1\over 2}I$ (expand it out - say, take $i=3$ - and you'll see why). If you think a bit about this example, you can derive the general formula for a finite geometric series, and with a bit of ingenuity even the formula for infinite geometric series!
