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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.

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    $\begingroup$ Welcome to Math Stack Exchange. Note $\frac 1 {2^j}=\left(\frac1 2\right)^j$ $\endgroup$ – J. W. Tanner Mar 25 '19 at 3:36
  • $\begingroup$ Is $i$ some finite number? Or are you trying to find the value of the series for any arbitrary $i$? $\endgroup$ – kkc Mar 25 '19 at 3:36
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    $\begingroup$ The reciprocals of each term of a geometric series is also a geometric one $\endgroup$ – lab bhattacharjee Mar 25 '19 at 3:37
  • $\begingroup$ i is finite but arbitrary. The sum does not approach infinity. $\endgroup$ – bpryan Mar 25 '19 at 3:39
  • $\begingroup$ 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 $\endgroup$ – J. W. Tanner Mar 25 '19 at 3:49
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It is a geometric series:

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

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  • $\begingroup$ harmonic series would have consecutive numbers in the denominators $\endgroup$ – J. W. Tanner Mar 25 '19 at 3:55
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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.

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