find the interval of convergence of the power series 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 

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:
$$\boxed{1 - \sqrt3 < x < 1+\sqrt3}$$
