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Proof of Divergence?

  1. Jul 10, 2011 #1
    1. The problem statement, all variables and given/known data

    Decide whether the series below is absolutely convergent, conditionally convergent, or divergent:

    [tex]\sum_{1}^{\infty}(2n+3)!/(n!)^2[/tex]

    3. The attempt at a solution

    By graphing the equation, I am confident that the series is divergent, but I don't know how to prove it. I cannot do the algebraic manipulation for a ratio test, assuming it is even possible, and none of the other tests seem applicable. Since it's apparently going to be divergent, I can't to a comparison test.

    That leaves either a straight limit test, or a limit comparison test. Unfortunately, it looks to me like the limit converges to zero. Factorial is stronger than the power function, but how can I prove factorial squared is weaker than a larger factorial in the numerator?

    Thanks :)
     
  2. jcsd
  3. Jul 10, 2011 #2
    The series diverges by the ratio test. Can you show what you've done using the ratio test?
     
  4. Jul 10, 2011 #3
    I can try....

    [tex]\lim_{n\rightarrow\infty}(2n+5)!/((n+1)!)^2 * (n!)^2/(2n+3)![/tex]

    If we expand and simplify [(n+1)!]^2, we get n!(n+1)n!(n+1)

    Cancel the two n! in the numerator and denominator to get....

    [tex]\lim_{n\rightarrow\infty}(2n+5)!/[(n+1)^2(2n+3)!][/tex]

    I can't further simplify this expression, or pove that it is greater than 1.

    Is this the right track?
     
  5. Jul 10, 2011 #4
    So far so good.

    Hint: [itex](2n+5)! = (2n+5)(2n+4)(2n+3)![/itex]
     
  6. Jul 10, 2011 #5
    Yeah I'm dumb...

    So canceling the common factorials gives us....

    [tex]\lim_{n\rightarrow\infty}(2n+5)(2n+4)/(n+1)(n+1)[/tex]

    Expand and approximate using leading coefficients to get...

    [tex]\approx 4n^2/n^2[/tex]

    Which converges to 4/1 > 1 at the limit, so divergent. I think this is correct now.

    Thank you so much!
     
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