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Math Help - Some problems with series

  1. #1
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    Some problems with series (Still need help)

    How do I determine the convergence/divergence of the following:

     <br />
\sum _{n=1}^{\infty }\left({\sqrt[n] 3}-1 \right)<br />

    I can't use limit comparison test with {\sqrt[n] 3} because it gives me 0. Root and ratio tests don't seem appropriate and I'm struggling to find a divergent series which is smaller than this to compare it with. Would it be possible to compare it with \sum _{n=1}^{\infty }\left(\ln\left(n \right)-1 \right)?

    As for

    \sum _{n=3}^{\infty }\frac{1}{\left( lnln\left(n \right) \right)^{\ln \left( n \right) }}

    I'm more or less completely stuck.

    Thanks for the help in advance.
    Last edited by Hweengee; November 12th 2008 at 08:55 AM.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Hweengee View Post
    How do I determine the convergence/divergence of the following:

     <br />
\sum _{n=1}^{\infty }\left({\sqrt[n] 3}-1 \right)<br />

    I can't use limit comparison test with {\sqrt[n] 3} because it gives me 0. Root and ratio tests don't seem appropriate and I'm struggling to find a divergent series which is smaller than this to compare it with.

    As for

    \sum _{n=3}^{\infty }\frac{1}{\left( lnln\left(n \right) \right)^{\ln \left( n \right) }}

    I'm more or less completely stuck.

    Thanks for the help in advance.
    For the first one, consider what power series \sqrt[n]{3} equals.

    For the second one

    Consider

    \frac{1}{\ln(\ln(n))^{\ln(n)}\leq{\frac{1}{\ln\lef  t(\frac{1}{n}\right)^{\ln(n)}

    That is a convergent alternating series
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  3. #3
    Junior Member toraj58's Avatar
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    Lightbulb

    i use Root Test to solve this problem:

    \lim_{n \to \infty} \sqrt[n]{\sqrt[n] {3} - 1}

    = \lim_{n \to \infty} \sqrt[n] {\sqrt[n] {3} (1- \frac {1}{\sqrt[n] {3}})}

    = \lim_{n \to \infty} {\sqrt 3}\sqrt[n] {1- \frac {1}{\sqrt[n] {3}}} = 0

    because 0 < 1 then the mention series Converge.
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by toraj58 View Post
    i use Root Test to solve this problem:

    \lim_{n \to \infty} \sqrt[n]{\sqrt[n] {3} - 1}

    = \lim_{n \to \infty} \sqrt[n] {\sqrt[n] {3} (1- \frac {1}{\sqrt[n] {3}})}

    = \lim_{n \to \infty} {\sqrt 3}\sqrt[n] {1- \frac {1}{\sqrt[n] {3}}} = 0

    because 0 < 1 then the mention series Converge.
    You performed your limit wrong, the root test gives one, which is indeterminate.
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  5. #5
    Junior Member toraj58's Avatar
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    Question

    are you sure it gives 1, why?....i got 0!
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  6. #6
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    yea root test gives a value of one. I'm not really sure what the power series representation of \sqrt [n] 3 is, but I'll go look it up. Thanks.
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  7. #7
    Junior Member toraj58's Avatar
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    Question

    i don't understand how both of you got 1, please explain!
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  8. #8
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    Quote Originally Posted by toraj58 View Post
    i use Root Test to solve this problem:

    \lim_{n \to \infty} \sqrt[n]{\sqrt[n] {3} - 1}

    = \lim_{n \to \infty} \sqrt[n] {\sqrt[n] {3} (1- \frac {1}{\sqrt[n] {3}})}

    = \lim_{n \to \infty} {\sqrt 3}\sqrt[n] {1- \frac {1}{\sqrt[n] {3}}} = 0

    because 0 < 1 then the mention series Converge.
    the 2nd and 3rd step aren't equivalent.
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  9. #9
    Junior Member toraj58's Avatar
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    Exclamation

    = \lim_{n \to \infty} {3^{\frac {1}{n^2}}}\sqrt[n] {1- \frac {1}{\sqrt[n] {3}}} = 0

    sorry for mistake, still it gives 0 to me
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  10. #10
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     <br />
\lim_{n \to \infty} {3^{\frac {1}{n^2}}}\sqrt[n] {1- \frac {1}{\sqrt[n] {3}}}<br />

    \lim_{n \to \infty} {3^{\frac {1}{n^2}}}=1

    \lim_{n \to \infty} \sqrt[n] {1- \frac {1}{\sqrt[n] {3}}}=1

    So their product must also be 1. I don't get how you get 0.
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  11. #11
    Junior Member toraj58's Avatar
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    <br />
\lim_{n\to\infty}{\frac {1}{\sqrt[n] 3}} = \lim_{n\to\infty} \frac {1}{3^{\frac {1}{n}}} = \frac {1}{3^0} = \frac {1}{1} = 1<br />

    so

    <br />
\lim_{n \to \infty} \sqrt[n] {1- \frac {1}{\sqrt[n] {3}}}= \lim_{n \to \infty} \sqrt[n] {1 - 1} = 0
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  12. #12
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    yes, but {\frac {1}{n}} tends to 0 as n tends to infinity. so you have an indeterminate form of 0^0 which is in this case 1. anyway i believe this series is divergent.
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  13. #13
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    You can write \sqrt[n]{3}-1=e^{\frac{\ln 3}{n}}-1\sim_{n\to\infty} \frac{\ln 3}{n}
    (Because \frac{e^x-1}{x}\to_{x\to 0} \exp'(0)=1 or just (this is the same) because of the usual Taylor expansion of the exponential at zero: e^x=1+x+o(x))
    As a consequence, since \sum_n \frac{1}{n} diverges, and we deal with series with positive coefficients, you can conclude that your series diverges as well.
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  14. #14
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    I have another elementary proof:

    Using the identity a^n-1=(a-1)(1+a+\cdots+a^{n-1}), we have:
    2=3-1=\left(\sqrt[n]{3}\right)^n-1=(\sqrt[n]{3}-1)\left(1+\sqrt[n]{3}+\left(\sqrt[n]{3}\right)^2 +\cdots+\left(\sqrt[n]{3}\right)^{n-1}\right) \leq (\sqrt[n]{3}-1) 3n (since every of the n terms in the sum is less than 3), so that \sqrt[n]{3}-1\geq\frac{2}{3n}. However, \sum_n\frac{1}{n} diverges.
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  15. #15
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    Thanks! That helped alot.
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