1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: Series 4.

  1. Nov 18, 2007 #1
    1. The problem statement, all variables and given/known data
    Determine whether the series converges or diverges.

    [tex]\sum_{n=3}^{\infty}\frac{1}{\left(\ln (\ln n)\right)^{\ln n}}

    3. The attempt at a solution

    wait I'll try to solve
    Last edited: Nov 18, 2007
  2. jcsd
  3. Nov 18, 2007 #2
    I deleted wrong solution here.
    Last edited: Nov 18, 2007
  4. Nov 18, 2007 #3
    [tex]\frac{1}{\left(\ln (\ln n)\right)^{\ln n}}=\frac{1}{e^{\ln n\ln (\ln (\ln n))}}=\frac{1}{n^{\ln (\ln (\ln n))}}[/tex]

    for n enough big [tex]\ln (\ln (\ln n))>2[/tex].and we can say that series converges
    Last edited: Nov 18, 2007
  5. Nov 19, 2007 #4

    Gib Z

    User Avatar
    Homework Helper

    That seems the correct method, though in case you want to clarify it, prove that ln(ln(ln n)) > 2 and is increasing, and also show that you are comparing it to say, zeta(2).
  6. Nov 19, 2007 #5
    To prove that it's increasing is it enough?

    [tex]\left(\ln (\ln (\ln n))\right)'=\frac{1}{n\ln (\ln n)}>0[/tex]
  7. Nov 19, 2007 #6


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    you got the derivative wrong. think chain rule.

    and in case you're wondering zeta(s)=[itex]\sum[/itex]1/n^s, so zeta(2)=[itex]\sum[/itex]1/n². for a complete solution, you should specify that you're comparing the general term to that of zeta(2).
    Last edited: Nov 19, 2007
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook