Finding the radius of convergence of a power series

[R Letter]

Homework Statement

Σ(n=0 to ∞) ((20)(-1)^n(x^(3n))/8^(n+1)

Homework Equations

Ratio test for Power Series: ρ=lim(n->∞) a_(n+1)/a_n

The Attempt at a Solution

I tried the ratio test for Power Series and it went like this:

ρ=lim(n->∞) (|x|^(3n+1)*8^(n+1))/(|x|^(3n)*8^(n+2))
=20|x|/8 lim(n->∞) 1
=20|x|/8

20|x|/8<1
|x|<2/5

So radius of convergence=2/5. However when I input the power series into Mathematica, it says that the radius of convergence should be 2 (http://goo.gl/9nAHoS)

Where am I going wrong in my calculation?

 

[RJLiberator]

Hi there mate, I am going to attempt to help.

Here are some possible errors that I am looking into:

First, x^(3n) when you make it n+1 in the numerator of the ratio test, it should be x^(3n+3), correct?

Well, that might do it. Does this help?

 

[Ray Vickson]

Homework Statement

Σ(n=0 to ∞) ((20)(-1)^n(x^(3n))/8^(n+1)

Homework Equations

Ratio test for Power Series: ρ=lim(n->∞) a_(n+1)/a_n

The Attempt at a Solution

I tried the ratio test for Power Series and it went like this:

ρ=lim(n->∞) (|x|^(3n+1)*8^(n+1))/(|x|^(3n)*8^(n+2))
=20|x|/8 lim(n->∞) 1
=20|x|/8

20|x|/8<1
|x|<2/5

So radius of convergence=2/5.However when I input the power series into Mathematica, it says that the radius of convergence should be 2 (http://goo.gl/9nAHoS)

Where am I going wrong in my calculation?

Try setting ##x^3 = t## and finding the radius of ##t##-convergence for ##\sum_n c_n t^n##. Then translate those results into statements about ##x##. Alternatively, do it over again, but repair the algebraic errors you made. (Avoidance of such errors is the reason I suggested looking at ##t## instead of ##x##.)

 

[RJLiberator]

Also, one other thing --> In your solution you have a 20 sticking out, however, it seems to me that when you take the a^(n+1) term and the a^n term and divide them, the 20 on each will cancel leaving |x^3|/8 <1

 

[R Letter]

@RJLiberator and @Ray Vickson, thank you both for the help. Sloppy algebra seems to be my perpetual downfall.