Homework Help: Calculus II - Power Series

1. Aug 20, 2011

GreenPrint

1. The problem statement, all variables and given/known data

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

2. Relevant equations

3. 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

Last edited: Aug 20, 2011
2. Aug 20, 2011

GreenPrint

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

3. Aug 20, 2011

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?

4. Aug 20, 2011

vela

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

5. Aug 21, 2011

Ah thanks