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Homework Help: Is this a valid proof for n! >2^n for all n>3

  1. Sep 25, 2016 #1
    1. The problem statement, all variables and given/known data
    Show that n!>2n for all n>3.

    2. Relevant equations
    I will attempt to use induction.

    3. The attempt at a solution
    We want to show that n!>2n for all n>3.
    Consider the case when n=4.
    [tex] 4! = 24 > 2^4 =16.[/tex]

    We want to show by way of induction that if the inequality is true for some k greater than 4, it is true for k+1.

    Assume the inequality holds for k>4. Then,
    [tex] 2^k < k! [/tex]
    [tex] 2*2^k < 2k![/tex]
    [tex]2^{k+1} < 2k![/tex]
    for k>4.

    But 2k! < (k+1)! for all k>4. Therefore

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

    and so by induction it follows that n! > 2n for all n > 3.
  2. jcsd
  3. Sep 25, 2016 #2
    Proof is ok, but I'm going to be annoying:



  4. Sep 25, 2016 #3
    Thank you. That always helps.

    The reason I assumed it held for k>4: 4 is the very first integer greater than 3. I figured if I can establish by induction that it holds for 4, then 5, 6, 7,...k, k+1,... then it must hold for all numbers greater than 3. There is no upper bound on k. I did spin my wheels a bit at first since I'm used to the base case being k=1.

    As for the second part, I multiplied by 2 because that would allow me to easily get the k+1 case because the base for the power is 2, and xax = xa+1. (also I don't believe the way everything relates to the number 2 here is coincidental. I think there might be a deeper relation between the base of the power and when it is larger than the factorial).

    Because I multiplied both sides by a positive integer, so the inequality should remain. The multiplication was done so I can get the k+1 case for the exponent. I then kind of lucked into the next part. I knew I had to get to some point where the left was less than the target quantity, and this worked out well.

    [tex]\lim_{n\rightarrow ∞ } \frac{2n!}{(n+1)!} = 0 [/tex]
    *EDIT I had n approaching zero for some reason

    and as for k =3,2,1 and 0, I manually tested them. k>4 is overkill for this part. But I figured that in order remain consistent I'd stick with k >4, since it is necessary for comparing the exponential to the factorial.
    Last edited: Sep 25, 2016
  5. Sep 25, 2016 #4


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    This is wrong.
  6. Sep 25, 2016 #5


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    There is, however, a small but significant error in your original proof. This is not related to insufficient justification, but is a genuine flaw.

    Big hint: what happened to the case ##n = 5##? Where did you cover that?
  7. Sep 25, 2016 #6
    I wrote it wrong. It's supposed to be as n approaches infinity.
  8. Sep 25, 2016 #7
    That means that for all ##\varepsilon>0##, there is an ##N## such that if ##n>N## we have ##2n! <\varepsilon (n+1)!##.

    Your claim is that for all ##n>3## we have ##2n!<(n+1)!##. Those are very distinct statements.
  9. Sep 25, 2016 #8
    Oh I see what you're saying.

    For that part when I answered your why question by posting the limit I just meant that because I knew that limit was zero it led me to the idea of trying to set up an inequality where

    2k+1 < 2k! < (k+1)! for k>4.

    I did not actually give a mathematical reason haha. I will think on it and try to articulate a better one.
  10. Sep 25, 2016 #9

    Ray Vickson

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    If you mean ##2 \times 2^k < (2k)! ## then that does not follow from the previous line. If you mean ##2 \times 2^k < 2 \times k!## then that does follow.
  11. Sep 25, 2016 #10


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    PeroK's question has been ignored. It raises an important point!
    Last edited: Sep 25, 2016
  12. Sep 26, 2016 #11
    Yes that's what I meant. I figured a factorial was only grouped with other terms if there were parentheses, but then again it's not like a factorial is in PEMDAS so I would be ignorant of such rules.

    I post these questions at work and unfortunately ran out of time. I was thinking about it, though. Sadly I'm at a loss. More information below...

    Okay... let me see...

    I have manually tested 5 and it works. 5! = 120 >25 = 32.
    Likewise with my last step, 25 = 32 < 2(5!) = 240 < (5+1)! = 6! = 720.

    As for the induction process, I've never done it where the base case wasn't one. Did I do it wrong with my base case of 4?

    Since I included k greater than 4, doesn't that already include 5. I don't quite understand what you're getting at. Any other hints? :/

    EDIT- since we're here, I did find a different way to do it working the other way.

    [tex](k +1)! = (k + 1)k! > (k + 1)2^k[/tex]
    and since k > 4 then
    [tex](k + 1)! > 2^k * 2 [/tex]
    [tex](k + 1)! > 2^{k+1}[/tex]
  13. Sep 26, 2016 #12


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    What you did was:

    a) Showed it was true for ##n = 4##

    b) Showed that if it was true for ##k > 4## it was true for ##k+1##

    But, ##k > 4## starts at ##k =5##. So, you didn't cover the step ##P(k = 4) \ \Rightarrow \ \ P(k = 5)##.

    The inductive stage of your proof should have been for ##k \ge 4##.

    On the other issues, I think you have let the questions confuse you and over-elaborated your answers. Your proof depended only on (if ##a, b## are positive):

    ##a < b \ \Rightarrow \ 2a < 2b##


    ##k > 2 \ \Rightarrow \ 2a < ka##

    You must see this, but recognising these simple facts and stating them simply and mathematically is something that generally does not seem to come naturally to many people.

    On a final point, I would tend to write either ##2(k!)## or ##(2k)!## to avoid any ambiguity. Like you, I don't know what PEMDAS has to say about factorials.
  14. Sep 26, 2016 #13
    So essentially if I replace the > with a ≥ that would fix the issue?

    So I could have omitted this part?

    Yeah I think I'll do that from now on.

    I appreciate the help by the way.
  15. Sep 26, 2016 #14


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    Yes, but you should see that yourself.

    There was nothing you could have omitted. You just needed to justify what you were doing a little more.
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