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Math Help - Hard one.

  1. #1
    Member Miss's Avatar
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    Hard one.

    hi

    Test the series for convergence:
    \sum_{n=1}^{\infty} \left( 1 - \sqrt[n]{n} \right)

    am sure, it will be tested by using one of the comparison tests
    but i can't handle it
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  2. #2
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    Quote Originally Posted by Miss View Post
    hi

    Test the series for convergence:
    \sum_{n=1}^{\infty} \left( 1 - \sqrt[n]{n} \right)

    am sure, it will be tested by using one of the comparison tests
    but i can't handle it
    nvm. I thought it diverged by the nth-Term test because  n^{\frac{1}{n}} goes to 0, but it doesn't. It goes to 1; thus, 1 - 1 = 0. It might converge. :S
    Last edited by lilaziz1; April 8th 2010 at 10:37 AM.
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  3. #3
    Member Miss's Avatar
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    Quote Originally Posted by lilaziz1 View Post
    It diverges by the nth-Term test.

     \sqrt[n]{n} is the same as  n^{\frac{1}{n}}
    But 1 - \sqrt[n]{n} \rightarrow 0 as n \rightarrow \infty

    the nth term test failed here ..
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  4. #4
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    Quote Originally Posted by Miss View Post
    hi

    Test the series for convergence:
    \sum_{n=1}^{\infty} \left( 1 - \sqrt[n]{n} \right)

    am sure, it will be tested by using one of the comparison tests
    but i can't handle it
    I think I have an answer that works, but I'm not 100% sure.

    We know that \sum_{n=1}^{\infty} \left( \frac{1}{n} \right) diverges.

    I believe that, for n > 2, it can be shown that \sqrt[n]{n} - 1 > \frac{1}{n}. Thus \sum_{n=1}^{\infty} \left( 1 - \sqrt[n]{n} \right) diverges.

    I tried it as a guess, and so I worked backwards from the desired result:

    \sqrt[n]{n} - 1 > \frac{1}{n}

    \sqrt[n]{n} > 1 + \frac{1}{n}

    \sqrt[n]{n} > \frac{n+1}{n}

    n > \left( \frac{n+1}{n} \right)^n

    n > \frac{(n+1)^n}{n^n}

    n^{n+1} > (n+1)^n

    It seems like the last statement holds true for all n > 2, and the steps could simply be written in reverse order to prove what we wanted to prove.
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  5. #5
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    Quote Originally Posted by undefined View Post
    I believe that, for n > 2, it can be shown that \sqrt[n]{n} - 1 > \frac{1}{n}. Thus \sum_{n=1}^{\infty} \left( 1 - \sqrt[n]{n} \right) diverges.

    I tried it as a guess, and so I worked backwards from the desired result:

    \sqrt[n]{n} - 1 > \frac{1}{n}

    \sqrt[n]{n} > 1 + \frac{1}{n}

    \sqrt[n]{n} > \frac{n+1}{n}

    n > \left( \frac{n+1}{n} \right)^n
    To prove this last inequality, you can say that the right hand side \left(1+\frac 1n\right)^n converges toward e. In particular, it is bounded, and hence less than n for large n.

    Alternatively, from the beginning, one can write \sqrt[n]{n}-1=e^{\frac{\log n}{n}}-1\sim\frac{\log n}{n} so if you know the comparison test using asymptotic equivalence, you can conclude using \frac{\log n}{n}\geq \frac{1}{n} and the divergence of \sum_n \frac{1}{n}.
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  6. #6
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    Quote Originally Posted by Laurent View Post
    To prove this last inequality, you can say that the right hand side \left(1+\frac 1n\right)^n converges toward e. In particular, it is bounded, and hence less than n for large n.

    Alternatively, from the beginning, one can write \sqrt[n]{n}-1=e^{\frac{\log n}{n}}-1\sim\frac{\log n}{n} so if you know the comparison test using asymptotic equivalence, you can conclude using \frac{\log n}{n}\geq \frac{1}{n} and the divergence of \sum_n \frac{1}{n}.
    Thanks for your explanation. I should have noticed the connection with e!

    Regarding the asymptotic solution, would you mind explaining how you got e^{\frac{\log n}{n}}-1\sim\frac{\log n}{n} ? I can verify that the LHS divided by the RHS tends to 1 as n tends to infinity, but I don't know why.
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  7. #7
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    Quote Originally Posted by undefined View Post
    Regarding the asymptotic solution, would you mind explaining how you got e^{\frac{\log n}{n}}-1\sim\frac{\log n}{n} ? I can verify that the LHS divided by the RHS tends to 1 as n tends to infinity, but I don't know why.
    You agree that e^u-1\sim u as u\to0 (just another way of writing \frac{e^u-1}{u}\to 1, which is the derivative \exp'(0)=1). Furthermore, \frac{\log n}{n}\to 0 as n\to\infty. We may therefore compose limits and deduce e^{\frac{\log n}{n}}-1\sim\frac{\log n}{n} as n\to\infty. Here it is!
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  8. #8
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    Quote Originally Posted by Laurent View Post
    You agree that e^u-1\sim u as u\to0 (just another way of writing \frac{e^u-1}{u}\to 1, which is the derivative \exp'(0)=1). Furthermore, \frac{\log n}{n}\to 0 as n\to\infty. We may therefore compose limits and deduce e^{\frac{\log n}{n}}-1\sim\frac{\log n}{n} as n\to\infty. Here it is!
    Thank you! It's very clear now.
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