Interval of Convergence/radius of convergence

[loganblacke]

Homework Statement

Find the interval of convergence of the series.. Sum from n=1 to infinity of ((-5^n)(x^n))/(n^(1/10)).

Find the radius of convergence.

Homework Equations

Ratio Test -> Lim abs( (An+1)/(An)) as n goes to infinity

The Attempt at a Solution

I used the ratio test to get to --> Lim as x goes to infinity of -5x(n/(n+1))^(1/10). I'm lost after this point. My notes say to use l'hospital's rule for infinity/infinity, which leaves me with the absolute value of -5x. plug that into the -b<-5x<b . How do you determine the value of b?

Thanks

 

[Mark44]

Homework Statement

Find the interval of convergence of the series.. Sum from n=1 to infinity of ((-5^n)(x^n))/(n^(1/10)).

Find the radius of convergence.

Homework Equations

Ratio Test -> Lim abs( (An+1)/(An)) as n goes to infinity

The Attempt at a Solution

I used the ratio test to get to --> Lim as x goes to infinity of -5x(n/(n+1))^(1/10). I'm lost after this point. My notes say to use l'hospital's rule for infinity/infinity, which leaves me with the absolute value of -5x. plug that into the -b<-5x<b . How do you determine the value of b?

I didn't check your work, but assuming it's correct so far, if the ratio test gives you a value of |-5x|, for what values does the ratio test tell you that the series converges?

 

[loganblacke]

I'm not sure where the |-5x| fits in but according to my notes the series converges if L < 1 and diverges if L >1. All of the problems that I have done so far have been -1 < x < 1 but i have no idea where the -1 & 1 come from.

 

[Mark44]

Right. And you found that L = 5|x|. Putting this fact together with what you know about convergence and divergence using the ratio test tells you what?

 

[loganblacke]

OK so regardless of what L is equal to, the -b < x < b always start from b=1? In which case the series converges between -.2 and .2?? I don't understand the "b" part.

 

[Mark44]

Forget the b.

You found that the limit L was 5|x|. The ratio test says the series converges if L < 1 and diverges if L > 1, so your series converges if 5|x| < 1 <==> |x| < .2 <==> -.2 < x < .2. The series may or may not converge at one or both endpoints of this interval. You need to check them separately.