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Sequence homework problem help please

  1. Jul 30, 2007 #1
    1. The problem statement, all variables and given/known data
    Determine wheatehr the sequence diiverges or converges:


    2. Relevant equations
    [tex] a_{n}=\frac{(n+2)!}{n!}[/tex]



    3. The attempt at a solution
    I was gonna treat it using limits but the factorial is not defined for a function.
    How do I deal with this?
    Edit: sorry it is suppose to be: [tex] a_{n}=\frac{(n+2)!}{n!}[/tex] not[tex] a_{n}=\frac{(n+1)!}{n!}[/tex]
     
    Last edited: Jul 30, 2007
  2. jcsd
  3. Jul 30, 2007 #2
    Try rewriting (n+1)! in a different form. What do you notice about the relationship between (n+1)! and n!?
     
  4. Jul 30, 2007 #3
    Ok I am pretty sure the it will diverge because (n+1)! goes faster to infinity than n!

    rewrite like: [tex] a_{n}= 1+\frac{1}{n!}[/tex] ?
     
  5. Jul 30, 2007 #4
    Your rewritten version of the equation is not equal to the original equation. In fact, I believe the rewritten equation converges to 1 (do you see why?), thereby contradicting your statement above.

    Please rewrite the numerator again, keeping in mind the goal of trying to simplify the equation.
     
  6. Jul 30, 2007 #5
    can i say n! = n(n-1)! and sub that in?
     
  7. Jul 30, 2007 #6
    Yes! You are on the right track. If you try rewriting the numerator instead of the denominator, I think you will have more luck.
     
  8. Jul 30, 2007 #7
    but how would i rewrite the numerator?
     
  9. Jul 30, 2007 #8

    matt grime

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    So, you know that n!= n(n-1)!... why is it not clear to you that that is precisely the same as (n+1)!=(n+1)n!? Or (n+1)!=(n+2)*(n+1)! ?????
     
  10. Jul 30, 2007 #9
    how is n!= n(n-1)! the same as (n+1)!=(n+1)n!??
     
  11. Jul 30, 2007 #10
    what I meant was how did u get that?
     
  12. Jul 31, 2007 #11

    Dick

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    The formula is true for all n. If n!=n*(n-1)! then changing n->n+1 gives (n+1)!=(n+1)*n!. In this sense, it's the 'same formula'. You may wish to try this for n->n+2. Can you show (n+2)!=(n+2)*(n+1)*n! at least for large enough n?
     
  13. Jul 31, 2007 #12
    Oh ok, thanks dick for clarifying, I just didn't think of it like that.
    Indeed I worked it out and n->n+2 is (n+2)!=(n+2)*(n+1)*n!.
     
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