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anonymous

  • 5 years ago

does the series 1/(n^2*3^n) converge or diverge as n goes from 1 to infinty?? what test do you use??

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  1. Regan
    • 5 years ago
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    if we continue to sub larger values of n we find that the denominator heads to infinity which means this functions converges onto 0

  2. anonymous
    • 5 years ago
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    thats it? you can take the limit of it? ok then thanks!

  3. Regan
    • 5 years ago
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    we can understand the behaviour of the function

  4. anonymous
    • 5 years ago
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    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.

  5. anonymous
    • 5 years ago
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    Do you know what the ratio test is?

  6. anonymous
    • 5 years ago
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    you take the limit of a(n+1) divided by a(n)..?

  7. anonymous
    • 5 years ago
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    Yep as n goes to infinity. If the value of the limit is less than one, then it's absolutely convergent.

  8. anonymous
    • 5 years ago
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    ok thank you so much!

  9. anonymous
    • 5 years ago
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    The limit is equal to 1/3, and hence the series converges. I can do the steps if you like.

  10. anonymous
    • 5 years ago
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    i got it! but i have another question..how do you find the radius of convergence? what does that even mean?

  11. anonymous
    • 5 years ago
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    In power series, there is a radius of convergence. That means a domain within the radius at which the series will converge.

  12. anonymous
    • 5 years ago
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    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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