Converging or Diverging?

Question

I have to determine whether
Sum(n=1 to inf) of (2n)!/n^n is converging or not...

I thought of doing the ratio rule but..no clue on how to solve it

Answer 1

Well, let us compute the ratio between the (n+1)th term and the nth:

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

Simplifying:

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

Factoring out the n+1:

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

Pair up the terms as follows:

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

Rewriting the term on the right:

(2n+2) (2n+1)/(n+1) (1 - 1/(n+1))^n

Now, taking the limits of each term, we see that the limit of the ratios between successive terms are:

∞*2*e^(-1)

Which is:

∞

Which is not less than 1, so the series diverges.

Answer 2

Like answerer #1 said, diverges.

The intuitive idea lies in comparing (2n)! and n^n as n changes to (n+1). When you compare (2n)! to (2(n+1))! = (2n+2)! = (2n)! * (2n+1)* (2n+2) you pick up two factors, or a factor of approximately 4n^2. On the other hand going from n^n to (n+1)^(n+1) you pick up apprimately one extra factor: the base n is approximately the base n+1 then the power increases. (You can see the intuition in #1 proof where the e showed up as factor.) So this suggests the terms don't even go to 0. You can then use Stirling's formula to show that limit n-> 00 of (2n)!/n^n is not zero, so the series diverges.