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: Show by induction that (2n-1) /(2n) <= 1/(3n+1) for all n

  1. Oct 5, 2013 #1
    Show by induction that (2n-1)!!/(2n)!! <= 1/(3n+1) for all n

    1. The problem statement, all variables and given/known data

    Show by induction that [itex](\frac{1\cdot3\cdot5\cdot\cdot\cdot(2n-1)}{2\cdot4\cdot6\cdot\cdot\cdot2n})^2[/itex] [itex]\leq[/itex][itex]\frac{1}{3n+1}[/itex] for n=1,2,3, ...

    3. The attempt at a solution
    It might not really be relevant, but I've attached my work. I tried to subprove something by induction to help my principal induction, but I got something that was harder than the original.

    P.S. I tried using the Physics forums android app, but I couldn't figure out how to post.

    Attached Files:

    Last edited: Oct 5, 2013
  2. jcsd
  3. Oct 5, 2013 #2
    For n=1,2,3 you have 1/2 <= 1/4, 3/8 <= 1/7 and 5/16 <= 1/10 respectively
    are you sure that's the problem you're trying to solve?
  4. Oct 5, 2013 #3
    Thanks for your response. By "n=1,2,3,...", the problem statement means the natural numbers. Sorry for the confusion.
    Also, induction is mandatory.
  5. Oct 5, 2013 #4
    I understand that, but the statement isn't true for all natural numbers since it isn't true for the first at least 10 of them
  6. Oct 5, 2013 #5
    Of course, I missed something essential! The whole left side is squared. I edited the original post. Thanks for pointing that out.
  7. Oct 5, 2013 #6
    But it still isn't true that ##(3/8)^2 < 1/7 ##
  8. Oct 5, 2013 #7


    User Avatar
    Science Advisor
    Homework Helper

    (3/8)^2 is less than 1/7.
  9. Oct 6, 2013 #8
    Yiu'll need to prove that: [tex] \left( \frac{2n+1}{2n+2} \right) ^2 \leq \frac{3n+1}{3n+4} [/tex].

    The easiest way to do this is to write out the cross product.
  10. Oct 6, 2013 #9
    Yes, of course; I managed to square 8 and get 16.
  11. Oct 6, 2013 #10


    User Avatar
    Science Advisor
    Homework Helper

    In case it's not obvious, dirk_mec1's idea it that you can express ##a_{n+1} \le b_{n+1}## as ##a_{n} \frac{a_{n+1}}{a_n} \le b_{n} \frac{b_{n+1}}{b_n}##. You know ##a_n \le b_n## by your induction hypothesis. If the ratios follow the same pattern you are done.
  12. Oct 7, 2013 #11
    Indeed, that is exactly my idea and I've written it out to see it leading to succes :).
  13. Oct 7, 2013 #12
    Awesome guys, it worked! Thanks!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted