What is the radius and interval of convergence for the given power series?

[GreenPrint]

Homework Statement

Find the radius of convergence and the interval of convergence of the series sigma[n=1,inf] ((3x-2)^n/(n^2*3^n))

Homework Equations

The Attempt at a Solution

sigma[n=1,inf] ((3x-2)^n/(n^2*3^n))
I applied the Root Test
p=lim n->inf |(3x-2)^n/(n^2*3^n)|^(1/n) = lim n->inf |(3x-2)/(3n^(2/n))| = 0
So the series Converges but I'm lost as to how to come up with the radius of convergence or the interval of convergence

 

[GreenPrint]

My bad I did this completley wrong let me work it out some more

 

[GreenPrint]

p=lim n->inf |(3x-2)^n/(n^2*3^n)|^(1/n) = lim n->inf |(3x-2)/(3n^(2/n))| = |x-2/3|

sense for the root test 0<= p < 1
I have
0<= |x-2/3| <1
whose solution is
-1/3 < x < 5/3
which is my interval of convergence
So the radius of convergence is
(5/3-1/3)/2 = 2/3

does this look better?

 

[vela]

Looks okay except you made a small error calculating the radius of convergence. You should find it equal 1.

 

[GreenPrint]

Ah thanks