Need help with a series (radius, convergence)

dmitriylm
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Homework Statement


Find the series' radius and interval of convergence. What what value of x does the series converge absolutely, conditionally?

Sum (n=0 to infinity) (nx^n)/((4^n)((n^2) +1)))


Homework Equations





The Attempt at a Solution


Not quite sure where to start with this. I've been having trouble with series in this course because its not as straightforward as all the other problems.
 
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haven't tried it, but would start with a ratio test
 
I was taught to start these problems with the ratio test. Then when you take the limit as n approaches infinity you can factor out the x. If the limit is 0 the radius is infinity and if the limit is infinity the radius is 0. If it it a number (c) you can multiply it by the x you factored out and set up the inequality -1<CX<1. Then solve for x and find if it converges at the endpoints by putting them in the initial sum and using any convergence test you wish, this will give you the radius of convergence.
 
Doing the ratio test and checking for the limit I found the limit to be x/4. Where do I go from there?
 
-1<x/4<1 so to solve for x multiply everything by 4 which gives -4<x<4 so the interval is -4,4 but we don't know if it is a closed or opened interval. To determine this plug -4 in for x in the original series and find if it converges or diverges if it converges it will be a bracket on the -4 if it diverges it is a parenthesis then do the same for 4.
 
however i tried it real quick and got 4x not x/4. I could be wrong but it might be worth a double check.
 
and you will nee dto check the boundary individually as teh ration test does not apply for a ratio of 1
 
DCASH88 said:
however i tried it real quick and got 4x not x/4. I could be wrong but it might be worth a double check.
No, you are correct. The radius of convergence is 1/4, not 4.

The series will converge absolutely inside the radius of converge, -1/4< x< 1/4. It may converge absolutely, conditionally, or not converge at x= -1/4 and/or x=1/4. Those will have to be checked separately.
 
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