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Radius / Interval of Convergence (Power Series)

  • Thread starter student45
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  • #1
student45
I need help finding the radius & interval of convergence of the following power series:

The sum from n=0 to infinity of...

(2-(n)^(1/2)) * (x-1)^(3n)

I think the ratio test is supposed to work, but I narrow the limit of the test down to 1/|x-1|^3 < 1, and this doesn't make sense to me. Is there an easier way, or am I doing something wrong?

Thanks.
 

Answers and Replies

  • #2
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[tex] \sum_{n=0}^{\infty} (2-n^{\frac{1}{2}})(x-1)^{3n} [/tex]

Is this the question?
 
Last edited:
  • #3
student45
Yeah, that's the one. Is the ratio test the right approach? Is 1/|x-1|^3 < 1 the right result? I'm just not sure where to go from there, or even if that is right for the limit. Thanks a lot for your help.
 
  • #4
student45
Oh except that only the "n" is raised to 1/2. Not (2-n). The quantity is (2-n^.5)
 
  • #5
student45
[tex] \sum_{n=0}^{\infty} (2-n^{\frac{1}{2}})(x-1)^{3n} [/tex]

Now that's right. Interval/Radius of convergence is the issue. I just don't know what I'm doing wrong with the ratio test.

Thanks.
 
  • #6
1,235
1
Using the ratio test:

[tex] (x-1)^{3}\; \frac{2-(n+1)^{\frac{1}{2}}}{2-n^{\frac{1}{2}}} [/tex] As [tex] n\rightarrow \infty [/tex] we get [tex] |(x-1)^{3}| < 1 [/tex]
 
  • #7
student45
I appreciate it. Thanks!!
 

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