Results 1 to 5 of 5

Math Help - An easy problem

  1. #1
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641

    An easy problem

    Here is a neat problem I did yesterday.

    Show that

    \sum_{k=2}^{n}\frac{1}{\log_k(x)}=\frac{1}{\log_{n  !}(x)}


    (Pre-calc students cover your ears)

    Then use it to show that

    \sum_{k=2}^{\infty}\frac{1}{\log_k(x)} converges \forall{x}>1


    Not too hard if you see it, but neat I think .
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by Mathstud28 View Post
    Here is a neat problem I did yesterday.

    Show that

    \sum_{k=2}^{n}\frac{1}{\log_k(x)}=\frac{1}{\log_{n  !}(x)}
    trivial


    (Pre-calc students cover your ears)

    Then use it to show that

    \sum_{k=2}^{\infty}\frac{1}{\log_k(x)} converges \forall{x}>1
    Does not converge for any x \ne 1, since:

    \sum_{k=2}^{\infty}\frac{1}{\log_k(x)}=\sum_{k=2}^  {\infty} \frac{\ln(k)}{\ln(x)} =\frac{1}{\ln(x)}\sum_{k=2}^{\infty} \ln(k)

    Some numerical evidence:

    Code:
    >k=2:100;
    >x=2;
    >s=1/logbase(x,k);
    >
    >S100=sum(s)
          524.765 
    >
    >
    >k=2:1000;
    >x=2;
    >s=1/logbase(x,k);
    >
    >S1000=sum(s)
           8529.4 
    >
    >
    >
    >k=2:10000;
    >x=2;
    >s=1/logbase(x,k);
    >
    >S10000=sum(s)
           118458 
    >

    RonL
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by CaptainBlack View Post
    trivial




    Does not converge for any x \ne 1, since:

    \sum_{k=2}^{\infty}\frac{1}{\log_k(x)}=\sum_{k=2}^  {\infty} \frac{\ln(k)}{\ln(x)} =\frac{1}{\ln(x)}\sum_{k=2}^{\infty} \ln(k)

    Some numerical evidence:

    Code:
    >k=2:100;
    >x=2;
    >s=1/logbase(x,k);
    >
    >S100=sum(s)
          524.765 
    >
    >
    >k=2:1000;
    >x=2;
    >s=1/logbase(x,k);
    >
    >S1000=sum(s)
           8529.4 
    >
    >
    >
    >k=2:10000;
    >x=2;
    >s=1/logbase(x,k);
    >
    >S10000=sum(s)
           118458 
    >

    RonL
    It's not trivial for Pre-calc Students .

    And yes, it was a trick question


    I just said that since it equatls \log_x(n!)

    Then \text{As }n\to\infty\Rightarrow{\log_{x}(n!)\to\infty}
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by Mathstud28 View Post
    It's not trivial for Pre-calc Students .
    It is trivial in the sense that it requires routine application of what they should know about change of base for logs and how to add fractions.

    And yes, it was a trick question


    I just said that since it equatls \log_x(n!)

    Then \text{As }n\to\infty\Rightarrow{\log_{x}(n!)\to\infty}
    You said use the identity to show that the given series converges, well it does not converge, and it obviously does not without recourse to the identity.

    RonL
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by CaptainBlack View Post
    It is trivial in the sense that it requires routine application of what they should know about change of base for logs and how to add fractions.



    You said use the identity to show that the given series converges, well it does not converge, and it obviously does not without recourse to the identity.

    RonL
    Apparently you never gave your students "set up questions"?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: February 9th 2013, 09:59 PM
  2. easy log problem, did i get it right?
    Posted in the Algebra Forum
    Replies: 10
    Last Post: April 21st 2010, 03:27 AM
  3. An easy Problem(or is it?)
    Posted in the Algebra Forum
    Replies: 2
    Last Post: October 15th 2009, 05:38 AM
  4. Should be a easy problem...
    Posted in the Algebra Forum
    Replies: 3
    Last Post: November 16th 2008, 06:34 PM
  5. easy problem! please help!
    Posted in the Algebra Forum
    Replies: 6
    Last Post: April 1st 2008, 09:10 PM

Search Tags


/mathhelpforum @mathhelpforum