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 $$ \frac{x-1}{x^2-2x+2} $$
Best Answer
For any $\;x\;$ such that $\;|x-1|<1\iff -1<x-1<1\iff -2<x<0$, you get a geometric series:
$$\sum_{n=0}^\infty(-1)^n\left(x-1\right)^{2n+1}=(x-1)\sum_{n=0}^\infty\left(-(x-1)^2\right)^n=(x-1)\frac1{1+(x-1)^2}=$$
$$=\frac{x-1}{x^2-2x+2}$$