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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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    Yes. $\;\qquad$2010-12-20
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    @Ronnie: Language: It's incorrect so ask if the "convergence of the series" is a particular number. You can ask if the series *converges* to a particular number or not, or about "the convergence" of the series (in which case the possible answers are "yes, it converges" or "no, it does not converge").2010-12-20
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    I'll only add the additional note that though rearranging a series isn't always valid, a particular rearrangement gives the same answer as the straightforward computation: $(1+1/4+1/16+1/64+\dots)-(1/2)(1+1/4+1/16+1/64+\dots)=2/3$2010-12-20
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    It is true that if $a_0+a_1+a_2+\cdots$ and $b_0+b_1+b_2+\cdots$ converge, then $a_0+b_0+a_1+b_1+a_2+b_2+\cdots$ converges to $(a_0+a_1+a_2+\cdots)+(b_0+b_1+b_2+\cdots)$.2010-12-20
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    @J.M.: But since this series is absolutely convergent, rearranging is not a problem. (-:2010-12-20
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    @Arturo: True, but here absolute convergence isn't needed, just convergence of the subseries, because the subseries are not rearranged.2010-12-20
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    Also, can anyone help me explain why the series converges in words...would something like this work...this geometric series(or is it arithmetic? i always get confused) converges because in converging series, the formula 1/1-r would be between -1 and 1 and in this series, the formula ends to be 2/3. But my problem is can't converging series also converge to numbers outside of -1 and 1?2010-12-20
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    @Ronnie: $-\frac12$ is between -1 and 1... the convergence of an infinite geometric series places conditions on the common ratio, not on the result.2010-12-20
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    Geometric is correct, not arithmetic. (An arithmetic sequence in one in which there is a constant *difference* between successive terms.) $1/(1-r)$ does not need to be between $-1$ and $1$, but $r$ must satisfy $-1$r$ to converge. The reason follows from the identity I mentioned in a comment on my answer, along with the fact that $|r|\lt 1$ implies that $r^{n+1}$ converges to $0$ as $n$ goes to $\infty$. – 2010-12-20
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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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    thank you but in words how would i explain that the series converges?2010-12-20
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    What words are you looking for? Apparently you have already learned that geometric series converge. When you learned about this, was it explained to you in words? How about the words, "It is an infinite geometric series with common ratio $-1/2$ and first term $1$, therefore it converges to $1/(1+1/2)$ [by the theorem on geometric series you learned]. If you understand the mathematics involved, you should be able to use some words to help explain the steps. If you do not, could you please ask a more specific question about what you do not understand?2010-12-20
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    But in case it helps, note that identity $1+r+r^2+\cdots+r^n=\frac{1-r^{n+1}}{1-r}$ leads to the reason why the geometric series converges when $|r|\lt 1$.2010-12-20
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    @Ronnie: Considering this and your previous question: what exactly do you mean by "explaining in words"? Could you give an example of what sort of explanation you're expecting? I know that a number of people, including myself, would prefer a mathematical line-by-line explanation than an "explanation in words", since mathematical symbols (usually) say so much with so little.2010-12-20
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    likewise but i was asked to explain 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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    Why should rearranging give the same result? This is homework after all, and that needs to be justified too...2010-12-20
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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