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

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

Homework Statement



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, ...

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.
 

Attachments

Last edited:

Answers and Replies

  • #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?
 
  • #3
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Thanks for your response. By "n=1,2,3,...", the problem statement means the natural numbers. Sorry for the confusion.
Also, induction is mandatory.
 
  • #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
 
  • #5
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Of course, I missed something essential! The whole left side is squared. I edited the original post. Thanks for pointing that out.
 
  • #6
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But it still isn't true that ##(3/8)^2 < 1/7 ##
 
  • #7
Dick
Science Advisor
Homework Helper
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But it still isn't true that ##(3/8)^2 < 1/7 ##
(3/8)^2 is less than 1/7.
 
  • #8
761
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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.
 
  • #9
329
34
(3/8)^2 is less than 1/7.
Yes, of course; I managed to square 8 and get 16.
 
  • #10
Dick
Science Advisor
Homework Helper
26,258
618
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.
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.
 
  • #11
761
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Indeed, that is exactly my idea and I've written it out to see it leading to succes :).
 
  • #12
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Awesome guys, it worked! Thanks!
 

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