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I was given the following question:

Calculate $1-\frac12+\frac14-\frac18+\frac1{16}- \frac1{32}+\frac1{64}-\cdots -\frac1{512}+\cdots;$ and express the answer as a simple fraction.

My approach was to use the following formula: $\frac1{1-r}$ where $r$ is the common ratio. In the end I got $2/3$. Am I correct?

...Edit...

Also, how would i be able to explain this by words?

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    so would this be a good word explanation...this geometric series converges because it has a common ratio of -.5 which is between$-1$and 1. By knowing the common ratio, i was able to find the limit of this series by using the formula 1/1-r where$r$is the common ratio. When -.5 is placed in the formula for the common ratio, you get 1/1.5 or 2/3. So this series converges to 2/32010-12-20

3 Answers 3

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Yes, and your reason is correct. If you want a simple way to check your work on such things, you could use a computer algebra system like Mathematica, or simply Wolfram Alpha:

http://www.wolframalpha.com/input/?i=sum+%28-1%2F2%29%5En

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    likewise but i was asked to ex$p$lain in words2010-12-20
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Couldn't resist that the 'solution' that instantly came to mind was to observe that if x is the supplied series, then 2x = 2 - x, from which x = 2/3 follows easily. Also, x = 1 - 2 + 4 - 8 + 16 ... implies 2x + x = 1 implies x = 1/3. BZZZT!!!

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You could also have split it up into two sums. $(1+\frac{1}{4}+\frac{1}{16}+...)-(\frac{1}{2}+\frac{1}{8}+...)$ $(1+\frac{1}{4}+\frac{1}{16}+...)-\frac{1}{2}(1+\frac{1}{4}+\frac{1}{16}+...)$ $\frac{1}{2}(1+\frac{1}{4}+\frac{1}{16}+...)$

EDIT:

After reading Moron's comment, I decided to post the alternative way to get to my final equation. $(1-\frac{1}{2})+(\frac{1}{4}-\frac{1}{8})+...$ $\frac{1}{2}(1+\frac{1}{4}+\frac{1}{16}+...)$

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    Your update isn't any more justified. $(1-1)+(1-1)+\cdots = 0+0+\cdots$. (Of course both of your ways of rewriting the sum are correct, but they are not steps that would be valid in general.)2010-12-20