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Homework Help: Ratio Test for series convergence factoring problems

  1. May 4, 2010 #1
    1. The problem statement, all variables and given/known data
    [tex]\Sigma[/tex]2nn!/(n+2)!


    2. Relevant equations
    I'm using the ratio test because there are factorials but I'm a little stuck on whether or not to factor out


    3. The attempt at a solution
    lim 2n+1(n+1)!/(n+3)!*(n+2)!/2n(n)! After I set it up here I'm not sure of how to factor out the factorials. would it end up looking like 2(n+1)(n+2)/(n+3) ?
     
  2. jcsd
  3. May 4, 2010 #2
    Close, it should simplify to [2(n+1)]/(n+3)
     
  4. May 4, 2010 #3
    Alright so the (n+2)! would factor out. I wasn't sure whether or not I would just factor out the factorial or since the (n+3)! is bigger it would factor out the (n+2) as well. Thanks for your help.
     
  5. May 4, 2010 #4
    (n+3)! factors out to (n+3)[(n+2)!]. You only have to factor it out that far since you want to cancel the (n+2)! in the numerator.

    Generally, you can factor any factorial the way I'll show you, but we'll use (n+3)! as an example: factor out like this, until you can stop to cancel out something in either the numerator or denominator depending on where the larger factorial is:

    ex.) (n+3)(n+2)(n+1)(n)(n-1)(n-2)...3(2)(1)
     
  6. May 4, 2010 #5

    Mark44

    Staff: Mentor

    You can make life easier on yourself by doing some simplification first, before applying any of the tests. n!/(n + 2)! = n!/[(n + 2)(n + 1)n!] = 1/[(n + 2)(n + 1)], so you series is identically equal to [itex]\sum 2^n/[(n + 2)(n + 1)][/itex],
     
  7. May 4, 2010 #6
    Aaaaahh Fantastic!! Thank you guys both so much you've been very helpful to a truly lost cause in Calculus.
     
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