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Calculus limit and sequence Question

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  1. Sep 13, 2015 #1
    1. The problem statement, all variables and given/known data
    prove: [tex] \lim_{n\rightarrow \infty} {\frac{n!}{2^n}}=\infty[/tex]

    2. Relevant equations
    Def. of a limit

    3. The attempt at a solution
    I would like to know if my solution is right or not. I think it is right but I would like to get a feedback. Please do not give me the answer, just directions/hints/things to think about, etc.

    I need to show that for every (let's assume) positive natural number K (because I am going positively large, so negative numbers are not reasonable), I need to show that the sequence is larger then K from some place.

    I thought to use the K I will be given to prove it. So I tried to find when:

    [tex] \frac{K!}{2^K}>K [/tex]

    and I have: [tex]\leftrightarrow K!>K*2^K \leftrightarrow (K-1)!>2^K [/tex]

    And this is true for all K>=6. The proof is by induction(please feel free to correct me if I am wrong).

    So, we take the max(K,7) and then it is true.

    What do you think? (This is just a draft, so it will be more formal).

    Thanks
    Thomas
     
  2. jcsd
  3. Sep 13, 2015 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    It is not sufficient to prove that the sequence exceeds every K for some element. Imagine the sequence 0, 1, 0, 2, 0, 3, 0, 4, ... - it will do the same, but it does not have infinity as limit.

    In addition to what you have shown, you can also show that the sequence is increasing monotonically. Alternatively, show that the elements of the sequence are larger than K not only for element n=K but for all beyond that (K+1, K+2, ...) as well
     
  4. Sep 13, 2015 #3
    Thank you.
    I don't know how I forgot about the monotonically of the sequence(proved it but left it out for some reason).

    If I knew how to show that the elements of the sequence are larger than K and beyond, I would prefer this solution, because I think it should be easier, but could not find the way to do this. Hints or ways to think about that would be great.

    Thank you again!
    Thomas
     
  5. Sep 13, 2015 #4

    mfb

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    2016 Award

    Staff: Mentor

    Well, once you know that the sequence is increasing monotonically and ##\frac{K!}{2^K}>K##, then ##\frac{n!}{2^n}>K## for n>K is a direct consequence.

    f(n)=n-10 as lower bound would work as well, the limit of this sequence is obvious.
     
  6. Sep 13, 2015 #5
    I don't know what kind of proof they are looking for, but you could just argue by which value is greater, the limit of the numerator as n->inf or the limit of the denominator as n->inf
     
  7. Sep 13, 2015 #6

    Mark44

    Staff: Mentor

    No, that's not valid. Both limits are infinite.
    $$\lim_{n \to \infty} n! = \infty$$
    $$\lim_{n \to \infty} 2^n = \infty$$
     
  8. Sep 13, 2015 #7
    oh sorry, it's been a while since calc 1, but I thought there was something that since the numerator approaches infinity faster than the denominator that the limit would tend to inf instead of zero?
     
  9. Sep 13, 2015 #8

    Mark44

    Staff: Mentor

    Since the OP needs to prove that the limit of the fraction is ∞, the numerator would have to get large more quickly than the denominator, but taking the limits of the numerator and denominator aren't necessarily helpful, and especially so in this case.
     
  10. Sep 14, 2015 #9
    Thank you all!

    In the end I have decided to also show that the sequence is increasing monotonically.

    Thank you all again for helping me.

    Thomas.
     
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