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Math Help - What if a ratio test of series = 1

  1. #1
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    What if a ratio test of series = 1

    SUM, from 1 to infinity of:

    ln(n)
    ________
    (n+1)^3

    I work this out using the ratio test for convergence and the answer I get is "1"...

    Does this sound right?
    If so, this then means that there are no other tests that can be done correct? The test is simply inconclusive.
    Or:
    Is there another test that should be performed next?
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  2. #2
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    Quote Originally Posted by orendacl View Post
    SUM, from 1 to infinity of:

    ln(n)
    ________
    (n+1)^3

    I work this out using the ratio test for convergence and the answer I get is "1"...

    Does this sound right?
    If so, this then means that there are no other tests that can be done correct? The test is simply inconclusive.
    Or:
    Is there another test that should be performed next?
    You could try the root test...

    Root test - Wikipedia, the free encyclopedia
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  3. #3
    Super Member Failure's Avatar
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    Quote Originally Posted by orendacl View Post
    SUM, from 1 to infinity of:

    ln(n)
    ________
    (n+1)^3

    I work this out using the ratio test for convergence and the answer I get is "1"...

    Does this sound right?
    No, it doesn't. The ratio test acually tells you that this series converges absolutely:

    \frac{a_{n+1}}{a_n}=\frac{\ln(n+1)}{\ln (n)}\cdot\frac{n^3}{n^3+3n^2+3n+1}\rightarrow 0

    The first factor goes to 1 as n goes to \infty, since

    \frac{\ln(n+1)}{\ln (n)}=\frac{\ln(n)+\ln(1+1/n)}{\ln(n)}

    and the second factor goes to 0.


    If so, this then means that there are no other tests that can be done correct?
    No: if the ratio test fails, the root test might still be used to prove absolute convergence.
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  4. #4
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    Quote Originally Posted by Failure View Post
    No, it doesn't. The ratio test acually tells you that this series converges absolutely:

    \frac{a_{n+1}}{a_n}=\frac{\ln(n+1)}{\ln (n)}\cdot\frac{n^3}{n^3+3n^2+3n+1}\rightarrow 0

    The first factor goes to 1 as n goes to \infty, since

    \frac{\ln(n+1)}{\ln (n)}=\frac{\ln(n)+\ln(1+1/n)}{\ln(n)}

    and the second factor goes to 0.



    No: if the ratio test fails, the root test might still be used to prove absolute convergence.

    Doesn't it \frac{\ln(n+1)}{\ln (n)}\cdot\frac{n^3}{n^3+3n^2+3n+1} go to  1 ?

    We can use this inequality

     \ln{n} < n < n+1


    If a ratio test of series = 1 , it means nothing .
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  5. #5
    Super Member Failure's Avatar
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    Quote Originally Posted by simplependulum View Post
    Doesn't it \frac{\ln(n+1)}{\ln (n)}\cdot\frac{n^3}{n^3+3n^2+3n+1} go to  1 ?
    Oops, yes, you are right (obviously I didn't even take the time to read carefully enough what I had written myself).

    Maybe it was because I was already quite sure that the series converges. (Why? - Because I know that \sum_n\frac{1}{n^2} converges and can be used to bound the given series from above.)

    If a ratio test of series = 1 , it means nothing .
    Sure, in that case, the ratio test does not decide the question of convergence vs. divergence.
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  6. #6
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    Try using the fact that for n > 0 that

     <br />
\frac{\ln n}{(n+1)^3} < \frac{\ln n}{n^3} < \frac{n}{n^3} = \frac{1}{n^2}<br />

    (and read the previous post).
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  7. #7
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    OK,
    I said that initially the test was inconclusive as via Ratio = 1, then using the SUM fron n to infinity of 1/n^2 as being a bound for the problem which "is" convergent, thus this is a CONVERGENT series.
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