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Homework Help: Does series converge or Diverge?

  1. Jul 31, 2007 #1
    Does series converge or Diverge????

    1. The problem statement, all variables and given/known data
    ((2^n)/((n+1)!)) Bottom of Summation: n=0, top is infinity

    3. The attempt at a solution

    Replacing the n's with n+1, then inverting and multiplying the following:

    ((2^n * 2) / ((n+1)+1)n!)) * ((n+1) * n! ) / 2^n

    I got 2, with is greater than 1, so it diverges according to the Ratio Test.
    Is this correct?
  2. jcsd
  3. Jul 31, 2007 #2
    Incorrect, it should be 0.

    "((n+1)+1)n!))" should be (n+2)(n+1)*n! - which cancels with the (n+1)*n!, leaving 2/(n+2)
  4. Jul 31, 2007 #3
    Thanks Proton. That makes sense. I did not even think to add the two 1's together.
  5. Jul 31, 2007 #4
    Well. Just writing it out on paper and it does not make sense to me what you said Proton. Why would ((n+1)+1)n! equat (n+2)(n+1) * n!?? It seems that if anything it would be just (n+2)*n!

    Then the result would be (2n+1)/n+2
  6. Jul 31, 2007 #5


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    2^(n+1)/2^n=2. ((n+1)+1)!/(n+1)!=(n+2)!/(n+1)!=(n+2)*(n+1)!/(n+1)!=n+2.
    So the total ratio is 2/(n+2)->0 as Proton said.
  7. Aug 1, 2007 #6
    Could you step me through this. It is just not making any sense.

    after substituting n+1 for all n's, I have:
    (2^n * 2)/((n+1)+1)n!)

    This equation to be multiplied by ((n+1)n!)/2^n

    I understand getting (n+2)n! after added the two ones together. Just don't see where that second n+1 comes from
  8. Aug 1, 2007 #7


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    Concentrate on substituting n->n+1 in (n+1)*n!. The first factor changes to (n+2) (I think you are ok on that) but the second factor needs to change to (n+1)! since it contains n and n should be changed to n+1. So the result is (n+1)n! -> (n+2)(n+1)!. Now split (n+1)!=(n+1)*n!.
  9. Aug 1, 2007 #8
    Ok.... i understand. I was not substituting the n+1 into the n!. I was cancelling the two n!'s. Thanks for your help. I understand now. Just simply overlooking that step.
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