Proving $\lim_{n \rightarrow \infty} \frac{n!}{n^n} = 0$ - Tutoring Maths Resits

[Zurtex]

I'm doing some tutoring tomorrow morning for a girl who needs to do some maths resits. I was looking through the papers and you need to prove that:

$$n^{n-1} \geq n! \quad \forall n > 1 \; \text{and} \; n \in \mathbb{N}$$

Which is fine, but it asks to you to do it by proving by definition that:

$$\lim_{n \rightarrow \infty} \frac{n!}{n^n} = 0$$

And I have to admit that I can't quite remember how to do this one, if someone could point me in the right direction that would be great, I remember seeing a proof for it so I'm sure it'll come back to me.

 

[Hurkyl]

Split it into a part that obviously goes to zero, and a part that is obviously bounded!

 

[iNCREDiBLE]

Stirling's formula/approximation works fine too.

 

[Zurtex]

Stirling's formula/approximation works fine too.

Too far ahead for the material she is expected to know and my brain isn't working at the moment, I don't know what you mean Hurkyl

 

[TD]

I doubt this would pass as a proof, but here 1/n goes to zero and the second limit has an equal number of factors in the nominator as denominator. The highest factors are equal (n) but in the nominator, they decrease while they don't in the denominator.

$$\mathop {\lim }\limits_{n \to \infty } \frac{{n!}}{{n^n }} = \mathop {\lim }\limits_{n \to \infty } \frac{{n!}}{{n \cdot n^{n - 1} }} = \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\mathop {\lim }\limits_{n \to \infty } \frac{{n!}}{{n^{n - 1} }} = \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\mathop {\lim }\limits_{n \to \infty } \frac{{\overbrace {n \cdot \left( {n - 1} \right) \cdot \ldots \cdot 2}^{n - 1}}}{{\underbrace {n \cdot n \cdot \ldots \cdot n}_{n - 1}}}=0$$

 

[George Jones]

$$\left( \frac{n \cdot (n-1) ... 2} {n \cdot n ... n} \right) \frac{1} {n}$$

The stuff in the big brackets is less greater than zero and less than 1 for n > 2.

Regards,
George