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Find the radius of convergence and the sum for the power series.

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

I used the ratio test to find the R.

$$\frac{(-1)^{n+1} (x-1)^{2(n+1)+1}}{(-1)^n (x-1)^{2n}}= $$

$$ =(-x+1)^3 $$

R = 1 (convergence radius)

The radius should be correct (let me know if I did something wrong). However I have a hard time finding the sum for the power series. I think I should write it as a geometric series.

Would that be something like this?

$$ (-1)^n (1-\frac{1}{x})x^{2n+1} $$

And if this is right how would I proceed. I tried the formula for geometric sequences which did not work. I would like to know how to solve this problem and maybe a general approach.

The answer should be $$ \frac{x-1}{x^2-2x+2} $$

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    $\begingroup$ a) Yes, the radius is $1$, hence we say $R=1$, not $x=1$. $x=1$ would be the center of convergence. b) $1-\frac1x$ is a constant for all $n$... $\endgroup$ Jun 9, 2017 at 21:43

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For any $\;x\;$ such that $\;|x-1|<1\iff -1<x-1<1\iff -2<x<0$, you get a geometric series:

$$\sum_{n=0}^\infty(-1)^n\left(x-1\right)^{2n+1}=(x-1)\sum_{n=0}^\infty\left(-(x-1)^2\right)^n=(x-1)\frac1{1+(x-1)^2}=$$

$$=\frac{x-1}{x^2-2x+2}$$

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  • $\begingroup$ Mathematica confirms DonAntonio's answer. $\endgroup$ Jun 9, 2017 at 21:50
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    $\begingroup$ I'd say mathematics confirms it...and this is pretty elementary stuff. $\endgroup$
    – DonAntonio
    Jun 9, 2017 at 21:56
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    $\begingroup$ Yep... fair enough. $\endgroup$ Jun 9, 2017 at 21:57

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