# Squeeze Theorem with limits n!/n as n approaches 0

## Homework Statement

The question asks to use the squeeze theorem to show that the limit of n!/n^n equals 0 as n approaches ∞.

## Homework Equations

I need to use the squeeze theorem to solve this problem, and I'm not sure what the upper limit is.

## The Attempt at a Solution

I found the lower limit to be 1/n^n, and I don't remember how I got here. I need some help.

Isn't it just n^1? Which is just n?

Dick
Homework Helper
You can write n!/n^n as (1/n)*(2/n)*(3/n)*...*((n-1)/n)*(n/n), right? All of those factors are less than or equal to 1. How many of those factors are less than or equal to 1/2?

hmm, isn't 2/2^2 equal to 1/2? So only one of those is less than or equal to 1/2?

Dick
Homework Helper
hmm, isn't 2/2^2 equal to 1/2? So only one of those is less than or equal to 1/2?

No, take n=6. That's 6!/6^6=(1/6)*(2/6)*(3/6)*(4/6)*(5/6)*(6/6). Isn't it? I think three of those are less than or equal to 1/2. The first three. Can you generalize? n odd is a little different, but don't worry about that right now.

okay so, for n=6. everything less than 3/6 is less than or equeal to one half? I'm not really understanding what you mean by generalize.

Dick
Homework Helper
okay so, for n=6. everything less than 3/6 is less than or equeal to one half? I'm not really understanding what you mean by generalize.

Generalize means how many for n=10, n=20 etc. Now how many for a general even value of n?

So for about half of the values of n, half of them will be great than or equal to 1/2, right?

Can you show n!/(nn) ≤ (1/n) ?

Dick