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Math Help - Series Comparison test

  1. #1
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    Series Comparison test

    Hi, just wondering what can i compare: \sum {\frac{1} {(ln (k))^9}} with to show it diverges?

    Also, for \sum { \frac{1} {k^{1 + \frac{1} {k}}} }, how would i apprach this to test divergence/convergence?

    Thanks in advance.
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  2. #2
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    Quote Originally Posted by shinn View Post
    [snip]
    Also, for \sum { \frac{1} {k^{1 + \frac{1} {k}}} }, how would i apprach this to test divergence/convergence?

    Thanks in advance.
    k^{1 + \frac{1}{k}} = k \cdot k^{\frac{1}{k}}. Now consider \lim_{k \rightarrow \infty} k^{1 + \frac{1}{k}} = \lim_{k \rightarrow \infty} \left( k \cdot k^{\frac{1}{k}}\right) = \left( \lim_{k \rightarrow \infty} k\right) \cdot \left( \lim_{k \rightarrow \infty} k^{\frac{1}{k}} \right) and think about the necessary condition for the convergence of a series.
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  3. #3
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    Would this be correct?

    Consider the inequality  \frac{1} {k^{1 + \frac{1}{k}}} < \frac{1}{k} as  0 < \frac{1}{k} < 1.

    But by the p-series test,  \sum { \frac{1}{k}} diverges.
    Hence, by the comparison test,  \sum { \frac{1} {k^{1 + \frac{1} {k}}} } must also diverge.
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  4. #4
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    Quote Originally Posted by shinn View Post
    Would this be correct?

    Consider the inequality  \frac{1} {k^{1 + \frac{1}{k}}} < \frac{1}{k} as  0 < \frac{1}{k} < 1.

    But by the p-series test,  \sum { \frac{1}{k}} diverges.
    Hence, by the comparison test,  \sum { \frac{1} {k^{1 + \frac{1} {k}}} } must also diverge.
    Your argument is not correct.

    To prove \sum \frac{1} {k^{1 + \frac{1}{k}}} is divergent you require  \frac{1} {k^{1 + \frac{1}{k}}} {\color{red}>} f(k) where \sum f(k) is divergent.

    Re-examine my previous post.
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  5. #5
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    Hi,thanks for your help so far.
    But, I don't really get the hint.
    I know that a series diverges if you take the limit of the individual term to infinity and doesn't approach to zero.

    But,
    \lim_{k \rightarrow \infty} \frac{1} {k^{1 + \frac{1}{k}}} = 0
    so its inconclusive?

    Regarding my inequality, would it work if i place a fudge factor like:
    \frac{1}{2k}<   \frac{1} {k^{1 + \frac{1}{k}}}

    Also adding to my first question:
    What would be the "usual" comparison to compare something like \sum {\frac{1} {(ln (k))^p}} , for different values of p? .
    Last edited by shinn; September 12th 2009 at 02:46 AM.
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  6. #6
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    Quote Originally Posted by mr fantastic View Post
    k^{1 + \frac{1}{k}} = k \cdot k^{\frac{1}{k}}. Now consider \lim_{k \rightarrow \infty} k^{1 + \frac{1}{k}} = \lim_{k \rightarrow \infty} \left( k \cdot k^{\frac{1}{k}}\right) = \left( \lim_{k \rightarrow \infty} k\right) \cdot \left( \lim_{k \rightarrow \infty} k^{\frac{1}{k}} \right) and think about the necessary condition for the convergence of a series.
    Quote Originally Posted by shinn View Post
    Hi,thanks for your help so far.
    But, I don't really get the hint.
    I know that a series diverges if you take the limit of the individual term to infinity and doesn't approach to zero.
    But,

    \lim_{k \rightarrow \infty} \frac{1} {k^{1 + \frac{1}{k}}} = 0

    so its inconclusive?
    [snip]
    \lim_{k \rightarrow \infty} k^{\frac{1}{k}} = 1.

    \lim_{k \rightarrow \infty} k = + \infty.

    Now draw the obvious conclusion.
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  7. #7
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    Sorry for this replay.
    but please i want the answer of the first series ?
    (1)/([ln n]^9)
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  8. #8
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    Quote Originally Posted by TWiX View Post
    Sorry for this replay.
    but please i want the answer of the first series ?
    (1)/([ln n]^9)
    \frac{1}{(\ln n)^9} > \frac{1}{n} for all n > m where m is a large enough number. So a proof using the comparison test is an obvious approach.
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