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Math Help - Series...

  1. #1
    MHF Contributor Also sprach Zarathustra's Avatar
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    Series...

    I need to check if the next infinite series convergent:

    Sigma (k=2-->inf) 1/{ln(k)}^a (when a real positive number)


    Thank you.
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  2. #2
    MHF Contributor chisigma's Avatar
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    Is...

    \lim_{k \rightarrow \infty} \frac{\ln ^{a} k} {k}= 0

    ... no matter which is a, so that...

    Kind regards

    \chi \sigma
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  3. #3
    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by chisigma View Post
    Is...

    \lim_{k \rightarrow \infty} \frac{\ln ^{a} k} {k}= 0

    ... no matter which is a, so that...

    Kind regards

    \chi \sigma
    This is not helping me, I can't use the limit test:
    lim(n-->inf) an/bn =L when 0<L<inf thet u can say something about the series...
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  4. #4
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    Use simple comparsion. For all a>0~\&~n\ge 3
    \frac{1}{[log(n)]^a}\ge  \frac{1}{a^a\cdot n}.
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  5. #5
    MHF Contributor chisigma's Avatar
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    The expression \lim_{k \rightarrow \infty} \frac{\ln^{a} k}{k} = 0 means that exists a K so that for any k>K is \frac{1}{\ln^{a} k} > \frac{1}{k}. Now the series \sum_{k=2}^{\infty} \frac{1}{k} diverges so that the series \sum_{k=2}^{\infty} \frac{1}{\ln^{a} k} also diverges...

    Kind regards

    \chi \sigma
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  6. #6
    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by Plato View Post
    Use simple comparsion. For all a>0~\&~n\ge 3
    \frac{1}{[log(n)]^a}\ge \frac{1}{a^a\cdot n}.

    For Plato, how can I prove this?

    Thank you...
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  7. #7
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    \begin{array}{rcl}<br />
   {\log (n^{\frac{1}<br />
{a}} )} &  \leqslant  & {n^{\frac{1}<br />
{a}} }  \\<br />
   {\log (n)} &  \leqslant  & {an^{\frac{1}<br />
{a}} }  \\<br />
   {\left[ {\log (n)} \right]^a } &  \leqslant  & {a^a n}  \\<br />
   {\frac{1}<br />
{{\left[ {\log (n)} \right]^a }}} &  \geqslant  & {\frac{1}<br />
{{a^a n}}}  \\<br /> <br />
 \end{array}
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