How do I prove that the series converges or diverges?

Question

E { n! / (n^n) }

I HAVE to use comparison tests, and CANNOT use The Ratio Test. Please go through step by step, do NOT just give me an answer

Answer 1

{ n! / (n^n) }

Claim: This converges. Let's take a look at the denominator.

n >= n
n >= n - 1
n >= n - 2
n >= n - 3
.
.
.
n >= 1

Therefore,

n^n >= n(n - 1)(n - 2)(n - 3)...(2)(1).
n^n >= n!

This means that, if we take the reciprocals of both sides,

1/n^n <= 1/n! {Note the sign change; reciprocals of positive numbers work like that.}

Multiply n! both sides of the inequality,

n!/(n^n) <= n!/n!
n!/(n^n) <= 1

But
lim 1 = 1
n -> infinity
and, if

n!/(n^n) <= 1

and {1} converges, it follows that {n!/(n^n)} must converge too.

Edit: My mistake. I confused sequence with series. The answerer below is correct.

Answer 2

I don´t know why not to use the ratio test which is,in this case the most reasonable
a_(n+1)/a_n=( n+1)*n^n/(n+1)^(n+1)= 1/(1+1/n)^n ==> 1/e <1
so the series is convergent
Unless you can use the Stirling´s formula for an equivalent of n!
I don´t find a comparison test and this is more complicated
Seeing the former answer I would like to remind you that the series with term b_n= 1 is a divergent one
so n!/n^n< 1 does not prove anything