Results 1 to 7 of 7

Thread: Series...

  1. #1
    MHF Contributor Also sprach Zarathustra's Avatar
    Joined
    Dec 2009
    From
    Russia
    Posts
    1,506
    Thanks
    1

    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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    6
    Is...

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

    ... no matter which is $\displaystyle a$, so that...

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor Also sprach Zarathustra's Avatar
    Joined
    Dec 2009
    From
    Russia
    Posts
    1,506
    Thanks
    1
    Quote Originally Posted by chisigma View Post
    Is...

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

    ... no matter which is $\displaystyle a$, so that...

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
    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...
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,796
    Thanks
    2827
    Awards
    1
    Use simple comparsion. For all $\displaystyle a>0~\&~n\ge 3$
    $\displaystyle \frac{1}{[log(n)]^a}\ge \frac{1}{a^a\cdot n}$.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    6
    The expression $\displaystyle \lim_{k \rightarrow \infty} \frac{\ln^{a} k}{k} = 0$ means that exists a $\displaystyle K$ so that for any $\displaystyle k>K$ is $\displaystyle \frac{1}{\ln^{a} k} > \frac{1}{k}$. Now the series $\displaystyle \sum_{k=2}^{\infty} \frac{1}{k}$ diverges so that the series $\displaystyle \sum_{k=2}^{\infty} \frac{1}{\ln^{a} k}$ also diverges...

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor Also sprach Zarathustra's Avatar
    Joined
    Dec 2009
    From
    Russia
    Posts
    1,506
    Thanks
    1
    Quote Originally Posted by Plato View Post
    Use simple comparsion. For all $\displaystyle a>0~\&~n\ge 3$
    $\displaystyle \frac{1}{[log(n)]^a}\ge \frac{1}{a^a\cdot n}$.

    For Plato, how can I prove this?

    Thank you...
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,796
    Thanks
    2827
    Awards
    1
    $\displaystyle \begin{array}{rcl}
    {\log (n^{\frac{1}
    {a}} )} & \leqslant & {n^{\frac{1}
    {a}} } \\
    {\log (n)} & \leqslant & {an^{\frac{1}
    {a}} } \\
    {\left[ {\log (n)} \right]^a } & \leqslant & {a^a n} \\
    {\frac{1}
    {{\left[ {\log (n)} \right]^a }}} & \geqslant & {\frac{1}
    {{a^a n}}} \\

    \end{array} $
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 5
    Last Post: Oct 3rd 2011, 01:12 AM
  2. Replies: 3
    Last Post: Sep 29th 2010, 06:11 AM
  3. Replies: 0
    Last Post: Jan 26th 2010, 08:06 AM
  4. Replies: 2
    Last Post: Sep 16th 2009, 07:56 AM
  5. Replies: 1
    Last Post: May 5th 2008, 09:44 PM

Search Tags


/mathhelpforum @mathhelpforum