Find the radius of convergence and interval of convergence 
Seems like you are suppose to do the root test to come up with the answer. but the 2x-5 in the numerator concerns me. the (-5) part. The root test says that the series has to have positive terms. With the - 5 in there. It makes me confused.
Please help! 
 A: This is just a geometric serires with ratio $\frac{2x-5}{3}$. For it to converege, we must have
$$\left|\frac{2x-5}{3}\right|<1\Longrightarrow|2x-5|<3\Longrightarrow-3<2x-5<3\Longrightarrow2<2x<8\Longrightarrow1<x<4,$$
which comes also from the root test as you mentioned. Thus, we have that the convergence radius is $3/2$ and the interval is $(1,4)$.
A: If that stupid $2x-5$ is the problem, you better replace $2x-5$ with $y$ and now calculate for $$\sum_{n=1}^{\infty} \frac{y^n}{3^n}$$ 
radius  would then be $3$
i.e., you have convergence when $-3<y<3$ 
i.e., $-3<2x-5<3$ 
i.e., $2<2x<8$
i.e., $1<x<4$ 
I guess this would be helpful for other similar problems swell.. 
A: You can use the ratio test. You can just raise the whole thing to the 1/n th power. It's a shortcut. According to the test, the absolute value of (2x-5)/3 must be less than 1. So just set up the inequality and solve for x.
A: Write $(2x-5)^n/3^n=(2/3)^n/(x-5/2)^n$. Then the convergence is not disturbed by the change of the centre of the series. Use the $lim(a_{n}/a_{n+1})$ test
