# Sum of the inverse of a geometric series?

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.

• Welcome to Math Stack Exchange. Note $\frac 1 {2^j}=\left(\frac1 2\right)^j$ Commented Mar 25, 2019 at 3:36
• Is $i$ some finite number? Or are you trying to find the value of the series for any arbitrary $i$?
– kkc
Commented Mar 25, 2019 at 3:36
• The reciprocals of each term of a geometric series is also a geometric one Commented Mar 25, 2019 at 3:37
• i is finite but arbitrary. The sum does not approach infinity. Commented Mar 25, 2019 at 3:39
• My point of noting $\frac 1 {2^j}=\left(\frac 1 2\right)^j$ was not that $\frac 1 {2^j}$ was incorrect but rather that this is a geometric series Commented Mar 25, 2019 at 3:49

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