# Does series converge or Diverge?

Does series converge or Diverge????

## Homework Statement

((2^n)/((n+1)!)) Bottom of Summation: n=0, top is infinity

## 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?

## Answers and Replies

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)

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

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

Dick
Science Advisor
Homework Helper
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.

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

Dick
Science Advisor
Homework Helper
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!.

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.