# 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?

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
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