Find the interval of convergence.

  • Thread starter Sabricd
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  • #1
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Homework Statement



[tex]\sum[/tex] n^(1/2)*x^n where n=1 and goes to infinity.

Sorry, I'm new at this. I was kind of hoping to get help finding the interval of convergence.
After using the Ratio Test I got for an answer the absolute value of x. I know I have to prove whether or not the endpoints (-1,1) converge or not. Can I just treat it as an alternating series and apply its absolute value and test for divergence?
 

Answers and Replies

  • #2
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So basically, you'll need to check if the series

[tex]\sum{\sqrt{n}}~\text{and}~\sum{(-1)^n\sqrt{n}}[/tex]

converge. Does the sequence [tex]\sqrt{n}[/tex] converge to 0? What can you conclude from this?
 
  • #3
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Well if I take the limit of it and if the limit = 0 then it is convergent. If not, it is divergent, correct?
 
  • #4
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So I could basically take the absolute value and say that it diverges by the Test of Divergence...right?
 
  • #5
22,089
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You can only say, if it doesnt converge to 0, then our serie is divergent.
If it does converge to 0, then you have to apply another test.
 
  • #6
22,089
3,286
So I could basically take the absolute value and say that it diverges by the Test of Divergence...right?
right!
 
  • #7
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Thank you!
 
  • #8
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How about [tex]\sum[/tex] x^n/n^3 where n=1 and goes to infinity. I know the radius for convergence is 1. However, I have to prove whether the endpoints (-1, 1) converge. Hence, would the answer be no, it diverges since x does not have a limit?
 
  • #9
22,089
3,286
You'll need to see whether the series

[tex]\sum{\frac{1}{n^3}}~\text{and}~\sum{\frac{(-1)^n}{n^3}}[/tex]

These series will both converge (absolutely)...
 
  • #10
27
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Yes!! p series!!! :)
 

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