1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Bounds of log(n)

  1. Oct 31, 2009 #1
    1. The problem statement, all variables and given/known data

    Prove that for any integer n >= 2,

    1/2 + 1/3 + ... + 1/n <= log(n) <= 1 + 1/2 + 1/3 + ... + 1/(n-1)


    2. Relevant equations

    None


    3. The attempt at a solution

    I can see pictorally why the inequality holds true but despite numerous am struggling to make any real progress! Any hints or tips on how to get started would be very much appreciated!
     
  2. jcsd
  3. Oct 31, 2009 #2
  4. Oct 31, 2009 #3

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    What is the pictoral reason?
     
  5. Oct 31, 2009 #4
    The picture I had in mind was of that of log (n) with step functions (of the values in the inequalities) both above and below the graph drawn out by log (n).

    It is this that leads me to think the proof must involve the use of bounding step functions, but I cannot see how to begin.
     
  6. Nov 1, 2009 #5
    I have had a further look at this and think that I can adapt the proof of the fact that the limiting difference between the harmonic series and natural logarithm tending to the Euler constant to prove one side of the inequality.

    However, this still leaves the other side of the inequality unsolved and the fact that I am not sure this is the approach I should be taking!

    Any hints would be most appreciated...
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook