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I am trying to find a power series representation of

$$ \frac{1-x}{1+x} $$

My textbook does it by adding one and subtracting one on the numerator. I understand this method but was wondering if anyone could give me some input on doing it by the following method:

$$(1-x)\sum_{n=0}^\infty (-1)^{n}x^{n} $$

where the series is just the geometric series replaced by $-x$.

After multiplying out I get stuck here:

$$\sum_{n=0}^\infty (-1)^{n}x^{n} - \sum_{n=0}^\infty (-1)^{n}x^{n+1} $$

I'm not sure how I could progress or if this method is actually doable, any input would be appreciated thanks!

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    @Reveillark Did you not read all the way through the problem? – 2017-02-04
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    Yes! I understand there are other ways to do it much like the one done in my textbook, I was just hoping to get some input on the method I stated above and whether it's valid or not. – 2017-02-04

2 Answers 2

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Re-index:

$$\sum_{n=0}^\infty(-1)^nx^n-\sum_{n=0}^\infty(-1)^nx^{n+1}=\sum_{n=0}^\infty(-1)^nx^n+\sum_{n=0}^\infty(-1)^nx^n-1$$

$$=-1+2\sum_{n=0}^\infty(-1)^nx^n$$

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    You haven't done the reindexing right: the constant term should be $1$ not $2$. – 2017-02-04
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    This looks like just the answer I need! I'm just a bit confused on the second step and how you got the -x at the end? – 2017-02-04
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    @RobArthan I can't find what I did wrong. Could you explain? – 2017-02-04
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    @melm I took the $n=0$ term out of the second original sum. – 2017-02-04
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    Doesn't it look like this: $(1 - x + x^2 - x^3 + \ldots) - (x - x^2 + x^ 3 - \ldots) = 1 - 2x + 2x^2 - 2x^ 3 + \ldots$? – 2017-02-04
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    @SimplyBeautifulArt I see! Thank you very much I understand now – 2017-02-04
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    @melm Sorry, RobArthan caught a mistake. Indeed, I think looking at his comment may do you better. Now, I added one to the second original sum and then re-indexed everything to look better, and to preserve equality, I subtracted one as well. – 2017-02-04
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    @SimplyBeautifulArt could I ask why you added 1 to the second original sum? Is that one of the terms of the original sum? I was just a bit confused as the original sum has x to the power of (n+1) so I'm not sure where the 1 came from. Thanks! – 2017-02-04
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    @melm Note:$$1+\sum_{n=0}^\infty(-1)^{n+1}x^{n+1}=(-1)^{-1+1}x^{-1+1}+\sum_{n=0}^\infty(-1)^{n+1}x^{n+1}=\sum_{n=-1}^\infty(-1)^{n+1}x^{n+1}=\sum_{n=0}^\infty\dots$$ – 2017-02-04
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    @SimplyBeautifulArt We've only just learned this, I didn't know I could do that haha thank you! – 2017-02-04
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    @melm :D Well, hope you enjoy! – 2017-02-04
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I would prefer do the following $$ \begin{align} \frac{1-x}{1+x}&=-1+\frac{2}{1+x}\\ &=-1+2\sum_{n=0}^{+\infty}(-1)^nx^n\\ &=\left[-1+\sum_{n=0}^{+\infty}(-1)^nx^n\right]+\sum_{n=0}^{+\infty}(-1)^nx^n\\ &=\left[-1+(1-x+x^2-x^3+\dots)\right]+\sum_{n=0}^{+\infty}(-1)^nx^n\\ &=(-x+x^2-x^3+\dots)+\sum_{n=0}^{+\infty}(-1)^nx^n\\ &=-x(1-x+x^2-x^3+\dots)+\sum_{n=0}^{+\infty}(-1)^nx^n\\ &=-x\sum_{n=0}^{+\infty}(-1)^nx^n+\sum_{n=0}^{+\infty}(-1)^nx^n\\ &=(-x+1)\sum_{n=0}^{+\infty}(-1)^nx^n. & \end{align} $$