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2^n < (n+2)!

  1. Jul 9, 2014 #1
    1. The problem statement, all variables and given/known data


    prove by induction

    [tex] 2^n < (n+2)!; \forall n \ge 0[/tex]

    P(0)

    [tex] 2^0 < (0+2)! [/tex] [easy]

    P(k)

    [tex]2^k<(k+2)![/tex]
    [tex]= 2^k < (k+2)(k+1)k![/tex]

    P(k+1)

    [tex]2^{k+1} < (k+1+2)! [/tex]
    [tex]= 2^k \cdot 2 < (k+3)(k+2)(k+1)k![/tex]

    its pretty clear that

    [tex]2 < k+3[/tex]

    how do I show that though
     
    Last edited: Jul 9, 2014
  2. jcsd
  3. Jul 9, 2014 #2

    micromass

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    I would personally just say it's obvious. But if you want, you can prove it by induction. It all depends on how you defined ##<## and what your axioms are.
     
  4. Jul 9, 2014 #3

    HallsofIvy

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    I see you recognize that you cannot just write "[itex]2^{k+1}<((k+1)+ 2)![/itex]" since that is what you want to find- you wrote it only to remind yourself what you want to arrive at.

    Your "induction hypothesis" is that [itex]2^k< (k+ 2)![/itex]. Now write [itex]2^{k+1}= 2(2^k)< 2(k+2)![/itex].

    Now, use your "2< k+ 3" to get [itex]2^{k+1}< 2(k+2)!< (k+2)!(k+ 3)= (k+3)!= (k+1+2)![/itex].

    To prove that 2< k+ 3, start from the fact that, since [itex]0\le k[/itex], [itex]-1< k[/itex] and add 3 to both sides.
     
  5. Jul 9, 2014 #4

    I am taking this

    [itex]2^{k+1}= 2(2^k)< 2(k+2)![/itex]

    by hypothesis, adding the multiple 2 to each side doesn't change the inequality

    then prove 2<k+3

    and do this from the fact

    [tex]0 \le k[/tex] so -1<k ---> -1+3<k+3 = 2< K+3

    then I can say [itex] 2(2^k)< 2(k+2)! < (k+2)!(k+3)[/itex].
     
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