Trying to find the interval of convergence of this series: run into a problem

[skyturnred]

Homework Statement

The series:

\sum^{n=\infty}_{n=0} \frac{(-1)^{n+2}(x^{3}+8)^{n+1}}{n+1}

Homework Equations

The Attempt at a Solution

using the ratio test, I get the following:

|x^{3}+8|<1, but I know that the radius of convergence must be in the form:
|x-a|<b, where the degree of x is 1. I can't find a way to get the degree of x to 1! Can someone help me? Thanks

 

[LCKurtz]

Homework Statement

The series:

\sum^{n=\infty}_{n=0} \frac{(-1)^{n+2}(x^{3}+8)^{n+1}}{n+1}

Homework Equations

The Attempt at a Solution

using the ratio test, I get the following:

|x^{3}+8|<1, but I know that the radius of convergence must be in the form:
|x-a|<b, where the degree of x is 1.

That form is for a power series in $x$. You have a power series in $x^3+8$. Just figure out what $x$ give $|x^3+8|<1$, whether or not you get an interval.

 

[skyturnred]

That form is for a power series in $x$. You have a power series in $x^3+8$. Just figure out what $x$ give $|x^3+8|<1$, whether or not you get an interval.

When I solve for x I get that the interval of convergence is (-3,(-7)^{1/3}). So in order to solve |x^{3}+8|<1, x must fall in that interval. But what I don't get is how to get the radius of convergence from this..

 

[LCKurtz]

When I solve for x I get that the interval of convergence is (-3,(-7)^{1/3}). So in order to solve |x^{3}+8|<1, x must fall in that interval. But what I don't get is how to get the radius of convergence from this..

$(-9)^{-\frac 1 3}$ isn't -3. If $a$ is the center of your interval you should be able to express it in the form $|x-a|<b$. But I'm not sure whether or not I would call $b$ a radius of convergence in this type of problem. You might not even get an interval. For example a similar problem with squares instead of cubes might come out something like $4 < x^2 < 9$ giving $-3<x<-2$ and $2 < x < 3$. Then you don't have an "interval of convergence".