# Find the sum for the power series

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.

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

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

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

• 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 $n$... Jun 9, 2017 at 21:43

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

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

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

• Mathematica confirms DonAntonio's answer. Jun 9, 2017 at 21:50
• I'd say mathematics confirms it...and this is pretty elementary stuff. Jun 9, 2017 at 21:56
• Yep... fair enough. Jun 9, 2017 at 21:57
• Why is $-2<x<0$ is that a typo?
– user864806
Nov 25, 2022 at 12:56
• @Bestmat Yes, it is. It should be $\;-1<x<1\Longleftrightarrow 0<x<2\;$ .Thanks, editing now Nov 26, 2022 at 9:55