Can anyone help with radius of convergence?

Question

What is the radius of convergence of ∑(n=1 to ∞) of [(x^n)/(n^2)]

Do I use the ratio test with an=[1/(n^2)] or with an=[(x^n)/(n^2)]? What is the difference between the two?

My book an=[1/(n^2)] for another example. Can I actually just drop the x^n?

Thanks!

Answer 1

To do this problem, there are two steps: find the radius of convergence, and test what happens on the border:

Applying the ratio test, you get:

|a(n+1)/a(n)| = |x * n^2/(n+1)| ... as n-->∞, this becomes

|a(n+1)/a(n)| = |x|.

Remember that the ratio test gives you convergence for a ratio less than 1, divergence for a ratio greater than 1, and a failed test for a ratio equal to 1. Therefore, this test gives you convergence for everything on the interval (-1,1) and divergence for anything in (-∞,-1),(1,∞). Thus, the radius of convergence is 1.



However, the ratio test fails at x = 1 and x = -1. So, you need to check both those situations individually:

For x=1, you get ∑1/n^2 (i.e. you drop the x^n). This converges by the p-test or integral test.

For x=-1, you get ∑(-1)^n/n^2. This converges by the alternating series test.



So, your interval of convergence is [-1,1]. It is necessary to do the second round of tests in order to figure out whether or not to include the, um, "edges."

Answer 2

Let An = [(x^n)/(n^2), x is part of the problem, you cant just drop it...



by the ratio test:



lim x--> inf A(n+1)/An = [(x^(n+1)/((n+1)^2)]/[(x^n)/(n^2)]



= lim x--> inf abs( x * (n/n+1)^2)



= abs(x)



this only converges if < 1



so the interval of convergence is IxI < 1



so the radius of convergence is 1

Answer 3

enable us to think of the era of convergence is (-3,3) for a topic. to discover the radius of convergence: [3-(-3)]/2 = 3 discover the area of the era and divide it by 2; you get the radius of convergence.