Convergence of n!/n^n Sequence

[ehrenfest]

[SOLVED] proving a seqeunce converges

Homework Statement

Prove that the sequence \frac{n!}{n^n} converges to 0.

Homework Equations

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.

 

[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?

 

[ehrenfest]

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?

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.

 

[tiny-tim]

That's right!

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