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

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  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

    Dick

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    (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

    Dick

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    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!
     
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