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Calculus II - Power Series

  • Thread starter GreenPrint
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  • #1
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Homework Statement



Find the radius of convergence and the interval of convergence of the series sigma[n=1,inf] ((3x-2)^n/(n^2*3^n))

Homework Equations





The Attempt at a Solution



sigma[n=1,inf] ((3x-2)^n/(n^2*3^n))
I applied the Root Test
p=lim n->inf |(3x-2)^n/(n^2*3^n)|^(1/n) = lim n->inf |(3x-2)/(3n^(2/n))| = 0
So the series Converges but I'm lost as to how to come up with the radius of convergence or the interval of convergence
 
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Answers and Replies

  • #2
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My bad I did this completley wrong let me work it out some more
 
  • #3
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p=lim n->inf |(3x-2)^n/(n^2*3^n)|^(1/n) = lim n->inf |(3x-2)/(3n^(2/n))| = |x-2/3|

sense for the root test 0<= p < 1
I have
0<= |x-2/3| <1
whose solution is
-1/3 < x < 5/3
which is my interval of convergence
So the radius of convergence is
(5/3-1/3)/2 = 2/3

does this look better?
 
  • #4
vela
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Looks okay except you made a small error calculating the radius of convergence. You should find it equal 1.
 
  • #5
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Ah thanks
 

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