Finding radius of convergence of series ?

[SMA_01]

Homework Statement

How would I find the radius of convergence of this series?

f(x)=10/(1-3x)² is represented as a power series f(x)=$$\sum$$ from n=0 to $$\infty$$ C_nXⁿ

Homework Equations

The Attempt at a Solution

Okay so I tried deriving, using d/dx(1/1-3x)=3/(1-3x)² and ended up with $$\sum$$ (3x)ⁿ and I derived this series to get $$\sum$$ 3ⁿnX^n-1

I'm lost where to go from here or if I even did it right...how would I find the radius of convergence?

 

[HallsofIvy]

The standard way to find the radius of convergence of a power series is to use the "ratio test": the series $\sum a_n$ converges if [tex]\lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right|< 1[tex] Here, $a_n= 3^nnx^{n-1}$ and $a_{n+1}= 3^{n+1}(n+1)x^n$ so that the ratio is <br /> $$\left|\frac{a_{n+1}}{a_n}\right|= \frac{3^{n+1}(n+1)|x|^n}{3^nn|x|^{n-1}}= 3\frac{n+1}{n}|x|$$<br /> 1?<br /> What is the limit of that as x goes to infinity? For what values of x is that less than 1?<br /> <br /> One can, however, show that a power series will converge as long as there is "no reason not to"! The fraction $10/(1- 3x)^2$ only has a problem when the denominator is 0. That is, when 1- 3x= 0 or x= 1/3. There is "no reason not to converge" all the way from 0 to 1/3.[/tex][/tex]

 

[Dick]

That looks pretty good. I don't think you've been careful enough to quite get the series exactly correct. But that's not going to change the radius of convergence. Now use a ratio test to find the interval of convergence.

 

[SMA_01]

Thanks a lot! I didn't know I could use the ratio test form there.