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Math Help - please help with convergence divergence

  1. #1
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    please help with convergence divergence

    Please look over the following series to determine wheather it is convergent or divergent by specific test. If my answer is correct, please just say the number of the problem and yes, if it is wrong, I would live to have an explanation, but if it's too much work, evn to know that it is wrong will help.
    Thank you very, very much.

    1) series infinity, n=1 (n^2-6)/(n+4) by N-th term test
    My answer: lim of the n-th term does not exist, test failed

    2) series infinity, n=1, 5^n/3^n (geometric series)
    My answer: r=5/3 greater than 1, geometric series diverges.

    3) series infinity, n=1, 1/n^(3/4) (p-series test)
    My answer: p=3/4 less than 1, series diverges.

    4) series infinity, n=1 (n^2-2n)/3n^3+4n^2) by limit comparison
    My answer: both series a(n) and b(n) diverge based on b(n)

    5) series infinity, n=1, 2^n/(2n+4)! by ratio test
    My answer: lim 1/2n^2 =0 less than 1, converges

    6) series infinity, n=1, (n-2)/(3n+4) by alternating test
    My answer: diverges

    7) series infinity, n=1, 1/(3n^4+1) by comparison test
    My answer: converges
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  2. #2
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    6) series infinity, n=1, (n-2)/(3n+4) by alternating test ?

    My answer: diverges
    diverges because it fails the nth term test.

    The rest look ok to me.
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  3. #3
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    Thank you

    skeeter,

    Thank you a lot.
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  4. #4
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    Quote Originally Posted by oceanmd View Post
    Please look over the following series to determine wheather it is convergent or divergent by specific test. If my answer is correct, please just say the number of the problem and yes, if it is wrong, I would live to have an explanation, but if it's too much work, evn to know that it is wrong will help.
    Thank you very, very much.

    1) series infinity, n=1 (n^2-6)/(n+4) by N-th term test
    My answer: lim of the n-th term does not exist, test failed
    The test does not "fail". Because it does not have a limit of 0, the sequence diverges.

    2) series infinity, n=1, 5^n/3^n (geometric series)
    My answer: r=5/3 greater than 1, geometric series diverges.

    3) series infinity, n=1, 1/n^(3/4) (p-series test)
    My answer: p=3/4 less than 1, series diverges.

    4) series infinity, n=1 (n^2-2n)/3n^3+4n^2) by limit comparison
    My answer: both series a(n) and b(n) diverge based on b(n)
    What "a(n)" and "b(n)" are you talking about?

    5) series infinity, n=1, 2^n/(2n+4)! by ratio test
    My answer: lim 1/2n^2 =0 less than 1, converges

    6) series infinity, n=1, (n-2)/(3n+4) by alternating test
    My answer: diverges
    What "alternating test do you mean? If you mean "If a(n) alternates sign and a(n) decreases to 0, the the series converges" that doesn't apply here. It says nothing about what happens if a(n) is not alternating. As skeeter said, the series diverges because (n-2)/(3n+4) does not go to 0.

    7) series infinity, n=1, 1/(3n^4+1) by comparison test
    My answer: converges
    What did you compare it to?
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  5. #5
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    Thank you for your comments

    I would greatly appreciate more explanation.
    1. lim =n/1, which means lim does not exist. According to N-th term test, if lim does not equal to 0, then the series diverges, if lim = 0 then the test failed, series could eitherconverge or diverge.
    How can I conclude that the series diverges? Is it correct to say, because lim does not equal to 0, the series diverges, even when lim does not exist?

    4. a(n) is the original series, b(n) is a fraction discarded of all terms except those having the highest degree.

    7. I compare it to b(n), which is 1/3n^4
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