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Existence of N

  1. Feb 19, 2006 #1
    Is there an integer n such that

    [tex]\sqrt{n!+n}-\sqrt{n!} > 1[/tex]

    ?
     
  2. jcsd
  3. Feb 19, 2006 #2

    Hurkyl

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    What have you tried?

    *Sigh* If you really have no clue at all where to begin, try approximating things.
     
  4. Feb 20, 2006 #3

    arildno

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    Congregrate, consummate, conjugate, conflagrate and contemplate.
     
  5. Feb 20, 2006 #4
    You may find that the approximation

    [tex]\sqrt{n!+n} = \sqrt{n!}\sqrt{1+n/n!} = \sqrt{n!}\,\Big(1+\frac{1}{2(n-1)!}+\cdots\Big)[/tex]

    is useful.
     
  6. Feb 20, 2006 #5
    Your function appears to have an absolute maximum (for positive integers, anyway) somewhere between n=1 and n=3 (near n=2 perhaps). The value of the function at n=2 is [tex] 2- \sqrt2 [/tex] which is about 0.585786 <1. So I would say no. Unless n is negative. I haven't a clue how to calculate it in that case.

    -Dan
     
  7. Feb 20, 2006 #6
    Yes there is indeed a max < 1, after which it tends to 0, very, very rapidly.

    Too bad, I was hoping to disprove Andrica's conjecture.
     
  8. Feb 23, 2006 #7

    arildno

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    The simplest way to see this, is to first multiply with the conjugate expression:
    [tex]\sqrt{n!+n}-\sqrt{n!}=\frac{(\sqrt{n!+n}+\sqrt{n!})(\sqrt{n!+n}-\sqrt{n!})}{\sqrt{n!+n}+\sqrt{n!}}\leq\frac{n}{2\sqrt{n!}}[/tex]
    One can easily prove the proposition from this bound:
    For any n>3, we have
    [tex]\frac{n+1}{2\sqrt{(n+1)!}}=\frac{n}{2\sqrt{n!}}\frac{1+\frac{1}{n}}{\sqrt{n+1}}<\frac{n}{2\sqrt{n!}}\frac{2}{2}=\frac{n}{2\sqrt{n!}}[/tex]
    I.e, the sequence is decreasing.
     
    Last edited: Feb 23, 2006
  9. Feb 23, 2006 #8

    VietDao29

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    I am not very sure that I understand your post correctly, but do you mean:
    [tex]\sqrt{(n + 1)! + (n + 1)} - \sqrt{(n + 1)!} \leq \frac{n + 1}{2 \sqrt{(n + 1)!}} < \frac{n}{2 \sqrt{n!}} \geq \sqrt{n!+n}-\sqrt{n!}[/tex]. And you conclude that:
    [tex]\sqrt{(n + 1)! + (n + 1)} - \sqrt{(n + 1)!} \leq \sqrt{n! + n} - \sqrt{n!}[/tex].
    Is that reasonable?
    Am I missing something?
     
  10. Feb 23, 2006 #9

    arildno

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    Hmm, no.
    Multiplying out the numerator, we have the identity:
    [tex]\sqrt{n!+n}-\sqrt{n!}=\frac{n}{\sqrt{n!+n}+\sqrt{n!}}\leq\frac{n}{2\sqrt{n!}}[/tex]
    This holds for ANY n.
    Thus, if you can show that this bound is always less than 1 for any n, then it follows that your original difference expression must also be less than 1 for any n.
     
  11. Feb 23, 2006 #10

    VietDao29

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    Whoops, srry for misinterpreting your post. :redface:
    Should've read it more carefully. :blushing:
     
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