find the interval of convergence of the power series

Question

like the title said i have to find the interval of convergence of this power series :

$$\sum_{n=1}^\infty{ ((-1)^n *(x-1)^{2n-1})\over 3^n}$$

I applied the ratio test and i got something like this: $$\left|\frac{(-1)*(x-1)^2}3\right|$$ I know that I have to find the interval and then study the end points but I dont know how to do it. I'm stuck at this step

thanks in advance

Answer 1

The absolute value you got was correct. \begin{align*} \left|\frac{(-1)(x-1)^2}{3}\right| & = \left|\frac{(x-1)^2}{3}\right|\\ &= \frac13\left|(x-1)^2\right| <1 \end{align*} Now we have $$\left|(x-1)^2\right| < 3$$ $$1-\sqrt3 < x < 1+\sqrt3$$ Writing out some terms: $$\sum_{n=1}^\infty{(-1)^n \cdot (x-1)^{2n-1}\over 3^n} = -\frac{x-1}{3} + \frac{(x-1)^3}{9}-\frac{(x-1)^5}{27} + \cdots$$

Plugging in $1+\sqrt3$ for $x$: $$-\frac{\sqrt3}{3} + \frac{\sqrt{3}}{3}- \frac{\sqrt3}{3} + \cdots$$

Which is a geometric series with common ratio $-1$ so it diverges. If we plug in $1-\sqrt3$ for $x$ we get the series $$\frac{\sqrt3}{3} - \frac{\sqrt3}{3}+\frac{\sqrt{3}}{3} - \cdots$$ Which diverges by the same logic as above. There is no endpoint convergence, so our interval of convergence is: $$\boxed{1 - \sqrt3 < x < 1+\sqrt3}$$

Answer 2

By Ratio test we need $$\left|\dfrac{-(x-1)^2}3\right|<1$$

As for real $a,|-a^2|=a^2$

$$(x-1)^2<3\iff-\sqrt3\le x-1\le\sqrt3$$