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

    hi

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

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

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

    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 $\displaystyle 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; Apr 8th 2010 at 10:37 AM.
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  3. #3
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    Quote Originally Posted by lilaziz1 View Post
    It diverges by the nth-Term test.

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

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

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

    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 $\displaystyle \sum_{n=1}^{\infty} \left( \frac{1}{n} \right)$ diverges.

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

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

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

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

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

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

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

    $\displaystyle n^{n+1} > (n+1)^n$

    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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    Quote Originally Posted by undefined View Post
    I believe that, for n > 2, it can be shown that $\displaystyle \sqrt[n]{n} - 1 > \frac{1}{n}$. Thus $\displaystyle \sum_{n=1}^{\infty} \left( 1 - \sqrt[n]{n} \right)$ diverges.

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

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

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

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

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

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

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

    Regarding the asymptotic solution, would you mind explaining how you got $\displaystyle e^{\frac{\log n}{n}}-1\sim\frac{\log n}{n}$ ? 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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    Quote Originally Posted by undefined View Post
    Regarding the asymptotic solution, would you mind explaining how you got $\displaystyle e^{\frac{\log n}{n}}-1\sim\frac{\log n}{n}$ ? 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 $\displaystyle e^u-1\sim u$ as $\displaystyle u\to0$ (just another way of writing $\displaystyle \frac{e^u-1}{u}\to 1$, which is the derivative $\displaystyle \exp'(0)=1$). Furthermore, $\displaystyle \frac{\log n}{n}\to 0$ as $\displaystyle n\to\infty$. We may therefore compose limits and deduce $\displaystyle e^{\frac{\log n}{n}}-1\sim\frac{\log n}{n}$ as $\displaystyle n\to\infty$. Here it is!
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  8. #8
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    Quote Originally Posted by Laurent View Post
    You agree that $\displaystyle e^u-1\sim u$ as $\displaystyle u\to0$ (just another way of writing $\displaystyle \frac{e^u-1}{u}\to 1$, which is the derivative $\displaystyle \exp'(0)=1$). Furthermore, $\displaystyle \frac{\log n}{n}\to 0$ as $\displaystyle n\to\infty$. We may therefore compose limits and deduce $\displaystyle e^{\frac{\log n}{n}}-1\sim\frac{\log n}{n}$ as $\displaystyle n\to\infty$. Here it is!
    Thank you! It's very clear now.
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