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Proof by induction, ##(n!)^{2} \le (2n)!##.

  1. Mar 1, 2017 #1
    1. The problem statement, all variables and given/known data
    I need to prove by induction that ##(n!)^{2} \le (2n)!##. I'm pretty sure about my preliminary work, but I just need some suggestions for the end.

    2. Relevant equations
    It is well known from a theorem that if ##a \le b## and ##c \ge 0##, then ##ca \le cb##.

    3. The attempt at a solution
    ATTEMPT AT PROOF: For ##n=1##, we can see that ##(1!)^{2} \le (2\cdot 1)!## holds. Now, assume that ##(k!)^{2} \le (2k)!## holds for ##k \in N##. Then it must be shown that ##n=k+1## holds. For that,

    ##[(k+1)!]^{2} = (k+1)^{2}k!^{2}## and by assumption ##(k!)^{2} \le (2k)!## implies with the theorem above that

    ##(k+1)^{2}k!^{2} \le (k+1)^{2}(2k)!##

    And this is the point at which I don't know how to proceed. I need to conclude that this is in fact less than ##[2(k+1)]!##, but I am unsure how to do that. I would appreciate any direction.
     
  2. jcsd
  3. Mar 1, 2017 #2

    Ray Vickson

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    Are you absolutely required to use induction? I ask, because a direct proof (avoiding induction) is almost trivial.
     
  4. Mar 1, 2017 #3

    Orodruin

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    What is ##(2k)!/(2(k+1))!## ?
     
  5. Mar 1, 2017 #4
    I assume you mean to use ##(2(k+1))!/(2(k+1))!## in which case the expression ##\frac{(2k)!}{(2(k+1))!}## is equivalent to ##\frac{1}{2(k+1)(2k+1)}##, and so this reduces my inequality to

    ##(k+1)^{2}(k!)^{2} \le \frac{k+1}{2(2k+1)}\cdot (2(k+1))! \le (2(k+1))!## since ##\frac{k+1}{2(2k+1)} < 1## for any ##k\in N##. Is that what you had in mind?
     
  6. Mar 1, 2017 #5
    I would prefer to use induction simply for the sake of practice. It is not required.
     
  7. Mar 1, 2017 #6

    Orodruin

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    Yes.
     
  8. Mar 1, 2017 #7

    PeroK

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    If you expand the factorials, I think you'll see it fairly obvious.
     
  9. Mar 1, 2017 #8
    Yes. You are right, thank you.
     
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