anonymous
  • anonymous
does the series 1/(n^2*3^n) converge or diverge as n goes from 1 to infinty?? what test do you use??
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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regan
  • regan
if we continue to sub larger values of n we find that the denominator heads to infinity which means this functions converges onto 0
anonymous
  • anonymous
thats it? you can take the limit of it? ok then thanks!
regan
  • regan
we can understand the behaviour of the function

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anonymous
  • anonymous
Not really. If the limit as n goes to infinity is 0, that means there still a possibility to be convergent or divergent. An applicable test to this series is the ratio test.
anonymous
  • anonymous
Do you know what the ratio test is?
anonymous
  • anonymous
you take the limit of a(n+1) divided by a(n)..?
anonymous
  • anonymous
Yep as n goes to infinity. If the value of the limit is less than one, then it's absolutely convergent.
anonymous
  • anonymous
ok thank you so much!
anonymous
  • anonymous
The limit is equal to 1/3, and hence the series converges. I can do the steps if you like.
anonymous
  • anonymous
i got it! but i have another question..how do you find the radius of convergence? what does that even mean?
anonymous
  • anonymous
In power series, there is a radius of convergence. That means a domain within the radius at which the series will converge.
anonymous
  • anonymous
do we do the ratio test again and the limit we get is L and we just find 1/L and thats the radius of convergence??

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