# Proving a seqeunce converges

1. Mar 22, 2008

### ehrenfest

[SOLVED] proving a seqeunce converges

1. The problem statement, all variables and given/known data
Prove that the sequence $\frac{n!}{n^n}$ converges to 0.

2. Relevant equations

3. The attempt at a solution
Given $\epsilon > 0$, how do I find N? We know that the nth term is less than or equal to
$$\left(\frac{n-1}{n}\right)^{n-1}$$
but that really does not help.

2. Mar 22, 2008

### tiny-tim

… in a fight, always pick off the smallest ones …

Hi ehrenfest!

Hint: write it (1/n)(2/n)(3/n)…(n/n);

the early bits go down much faster than the later bits; so can you see a way of splitting off some of the early bits, and prove that they tend to zero?

3. Mar 22, 2008

### ehrenfest

Yes, I was trying to do something like that. If I can show that any factor in that product goes to 0, then the whole thing has to because the rest will be less than 1. If I collect the first n/2 ceiling terms, then I get (1/2)^(n/2) if n is even, and that needs to got to zero, doesn't it? I see, thanks. When n is odd we get something similar.

4. Mar 22, 2008

### tiny-tim

That's right!

(And don't forget, your proof should begin something like "For any epsilon, choose N such that 2^-N …")