# Homework Help: Does series converge or Diverge?

1. Jul 31, 2007

### bengaltiger14

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. Jul 31, 2007

### proton

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)

3. Jul 31, 2007

### bengaltiger14

Thanks Proton. That makes sense. I did not even think to add the two 1's together.

4. Jul 31, 2007

### bengaltiger14

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

5. Jul 31, 2007

### Dick

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.

6. Aug 1, 2007

### bengaltiger14

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

7. Aug 1, 2007

### Dick

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!.

8. Aug 1, 2007

### bengaltiger14

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.